I have solved the following problem, but I am not sure about the correctness of my solution so I would be grateful if someone would check it and give some feedback about it, thanks.
Suppose $f:\mathbb{R}^3\to\mathbb{R}$ is $\mathscr{C}^2$ and $\frac{\partial f}{\partial z}(\mathbf{a})\neq 0$, so that $f=0$ defines $z$ implicitly as a $\mathscr{C}^2$ function $\phi$ of $x$ and $y$ near $\bar{\mathbf{a}}.$ Show that $$\frac{\partial^2 \phi}{\partial x^2}(\bar{\mathbf{a}})=-\frac{\frac{\partial^2 f}{\partial z^2}\left(\frac{\partial f}{\partial x}\right)^2-2\frac{\partial^2 f}{\partial x\partial z}\frac{\partial f}{\partial x}\frac{\partial f}{\partial z}+\frac{\partial^2 f}{\partial x^2}\left(\frac{\partial f}{\partial z}\right)^2}{\left(\frac{\partial f}{\partial z}\right)^3}$$ where all the partial derivatives on the right-hand side are evaluated at $\mathbf{a}.$
In the following $\bar{\mathbf{x}}=\begin{pmatrix}x\\y\end{pmatrix}$ and $\mathbf{x}=\begin{pmatrix}x\\y\\z\end{pmatrix}$
My solution:
$D\phi(\bar{\mathbf{x}})=\begin{bmatrix}\frac{\partial\phi}{\partial x}(\bar{\mathbf{x}}) & \frac{\partial\phi}{\partial y}(\bar{\mathbf{x}})\end{bmatrix}=\begin{bmatrix}-\frac{\frac{\partial f}{\partial x}(\mathbf{x})}{\frac{\partial f}{\partial z}(\mathbf{x})} & -\frac{\frac{\partial f}{\partial y}(\mathbf{x})}{\frac{\partial f}{\partial z}(\mathbf{x})}\end{bmatrix}$.
Now, if we define $\mathbf{h}:\mathbb{R}^2\to\mathbb{R}^3, \mathbf{h}\begin{pmatrix}x\\y\end{pmatrix}= \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}x\\y\\\phi\begin{pmatrix}x\\y\end{pmatrix}\end{bmatrix}$, $g:\mathbb{R}^3\to\mathbb{R}, g\begin{pmatrix}x\\y\\z\end{pmatrix}=-\frac{\frac{\partial f}{\partial x}(\mathbf{x})}{\frac{\partial f}{\partial z}(\mathbf{x})}$ we have that $\frac{\partial\phi}{\partial x}=g\circ\mathbf{h}:\mathbb{R}^2\to\mathbb{R}$ so
\begin{align*} D\frac{\partial\phi}{\partial x}(\bar{\mathbf{x}}) &=\begin{bmatrix}\frac{\partial^2\phi}{\partial x^2}(\bar{\mathbf{x}}) & \frac{\partial^2\phi}{\partial y\partial x}(\bar{\mathbf{x}} )\end{bmatrix}=D(g\circ\mathbf{h})(\bar{\mathbf{x}})=Dg\left(\mathbf{h}(\bar{\mathbf{x}})\right)D\mathbf{h}(\bar{\mathbf{x}})\\ &=Dg(\mathbf{x})D\mathbf{h}(\bar{\mathbf{x}})=\begin{bmatrix}\frac{\partial g}{\partial x}(\mathbf{x}) & \frac{\partial g}{\partial y}(\mathbf{x}) & \frac{\partial g}{\partial z}(\mathbf{x})\end{bmatrix} \begin{bmatrix}\frac{\partial x}{\partial x}(\bar{\mathbf{x}}) & \frac{\partial x}{\partial y}(\bar{\mathbf{x}})\\ \frac{\partial y}{\partial x}(\bar{\mathbf{x}}) & \frac{\partial y}{\partial y}(\bar{\mathbf{x}})\\ \frac{\partial z}{\partial x}(\bar{\mathbf{x}}) & \frac{\partial z}{\partial y}(\bar{\mathbf{x}})\end{bmatrix}\\ &=\begin{bmatrix}\frac{\partial}{\partial x}\left(-\frac{\frac{\partial f}{\partial x}(\mathbf{x})}{\frac{\partial f}{\partial z}(\mathbf{x})}\right) & \frac{\partial}{\partial y}\left(-\frac{\frac{\partial f}{\partial x}(\mathbf{x})}{\frac{\partial f}{\partial z}(\mathbf{x})}\right) & \frac{\partial}{\partial z}\left(-\frac{\frac{\partial f}{\partial x}(\mathbf{x})}{\frac{\partial f}{\partial z}(\mathbf{x})}\right)\end{bmatrix}\begin{bmatrix}1 & 0\\ 0 & 1\\ \frac{\partial\phi}{\partial x}(\bar{\mathbf{x}}) & \frac{\partial\phi}{\partial y}(\bar{\mathbf{x}})\end{bmatrix}\\ &=\begin{bmatrix}\frac{\partial}{\partial x}\left(-\frac{\frac{\partial f}{\partial x}(\mathbf{x})}{\frac{\partial f}{\partial z}(\mathbf{x})}\right)+\frac{\partial}{\partial z}\left(-\frac{\frac{\partial f}{\partial x}(\mathbf{x})}{\frac{\partial f}{\partial z}(\mathbf{x})}\right)\cdot\frac{\partial\phi}{\partial x}(\bar{\mathbf{x}}) & \frac{\partial}{\partial y}\left(-\frac{\frac{\partial f}{\partial x}(\mathbf{x})}{\frac{\partial f}{\partial z}(\mathbf{x})}\right)+\frac{\partial}{\partial z}\left(-\frac{\frac{\partial f}{\partial x}(\mathbf{x})}{\frac{\partial f}{\partial z}(\mathbf{x})}\right)\cdot\frac{\partial\phi}{\partial y}(\bar{\mathbf{x}})\end{bmatrix} \end{align*}
It follows that \begin{align*} \frac{\partial^2\phi}{\partial x^2}(\bar{\mathbf{x}})&=\frac{\partial}{\partial x}\left(-\frac{\frac{\partial f}{\partial x}(\mathbf{x})}{\frac{\partial f}{\partial z}(\mathbf{x})}\right)+\frac{\partial}{\partial z}\left(-\frac{\frac{\partial f}{\partial x}(\mathbf{x})}{\frac{\partial f}{\partial z}(\mathbf{x})}\right)\cdot\frac{\partial\phi}{\partial x}(\bar{\mathbf{x}})\\ &=-\frac{\frac{\partial^2 f}{\partial x^2}(\mathbf{x})\frac{\partial f}{\partial z}(\mathbf{x})-\frac{\partial f}{\partial x}(\mathbf{x})\frac{\partial^2 f}{\partial x\partial z}(\mathbf{x})}{\left(\frac{\partial f}{\partial z}(\mathbf{x})\right)^2}-\frac{\frac{\partial^2 f}{\partial z\partial x}(\mathbf{x})\frac{\partial f}{\partial z}(\mathbf{x})-\frac{\partial f}{\partial x}(\mathbf{x})\frac{\partial^2 f}{\partial z^2}(\mathbf{x})}{\left(\frac{\partial f}{\partial z}(\mathbf{x})\right)^2}\cdot\left(-\frac{\frac{\partial f}{\partial x}(\mathbf{x})}{\frac{\partial f}{\partial z}(\mathbf{x})}\right)\\ &\overset{f\in\mathscr{C}^2}{=}-\frac{\frac{\partial^2 f}{\partial x^2}(\mathbf{x})\frac{\partial f}{\partial z}(\mathbf{x})-\frac{\partial f}{\partial x}(\mathbf{x})\frac{\partial^2 f}{\partial x\partial z}(\mathbf{x})}{\left(\frac{\partial f}{\partial z}(\mathbf{x})\right)^2}+\frac{\frac{\partial^2 f}{\partial x\partial z}(\mathbf{x})\frac{\partial f}{\partial x}(\mathbf{x})\frac{\partial f}{\partial z}(\mathbf{x})-\left(\frac{\partial f}{\partial x}(\mathbf{x})\right)^2\frac{\partial^2 f}{\partial z^2}(\mathbf{x})}{\left(\frac{\partial f}{\partial z}(\mathbf{x}) \right)^3}\\ &= \frac{\frac{\partial^2 f}{\partial x\partial z}(\mathbf{x})\frac{\partial f}{\partial x}(\mathbf{x})\frac{\partial f}{\partial z}(\mathbf{x})-\frac{\partial^2 f}{\partial x^2}(\mathbf{x})\left(\frac{\partial f}{\partial z}(\mathbf{x})\right)^2}{\left(\frac{\partial f}{\partial z}(\mathbf{x})\right)^3}+\frac{\frac{\partial^2 f}{\partial x\partial z}(\mathbf{x})\frac{\partial f}{\partial x}(\mathbf{x})\frac{\partial f}{\partial z}(\mathbf{x})-\left(\frac{\partial f}{\partial x}(\mathbf{x})\right)^2\frac{\partial^2 f}{\partial z^2}(\mathbf{x})}{\left(\frac{\partial f}{\partial z}(\mathbf{x}) \right)^3}\\ &=\frac{ -\frac{\partial^2 f}{\partial z^2}(\mathbf{x})\left(\frac{\partial f}{\partial x}(\mathbf{x})\right)^2 + 2 \frac{\partial^2 f}{\partial x\partial z}(\mathbf{x})\frac{\partial f}{\partial x}(\mathbf{x})\frac{\partial f}{\partial z}(\mathbf{x})-\frac{\partial^2 f}{\partial x^2}(\mathbf{x})\left(\frac{\partial f}{\partial z}(\mathbf{x})\right)^2}{\left(\frac{\partial f}{\partial z}(\mathbf{x})\right)^3}\\ &=-\frac{\frac{\partial^2 f}{\partial z^2}(\mathbf{x})\left(\frac{\partial f}{\partial x}(\mathbf{x})\right)^2 - 2 \frac{\partial^2 f}{\partial x\partial z}(\mathbf{x})\frac{\partial f}{\partial x}(\mathbf{x})\frac{\partial f}{\partial z}(\mathbf{x})+\frac{\partial^2 f}{\partial x^2}(\mathbf{x})\left(\frac{\partial f}{\partial z}(\mathbf{x})\right)^2}{\left(\frac{\partial f}{\partial z}(\mathbf{x})\right)^3} \end{align*}
thus $$\frac{\partial^2\phi}{\partial x^2}(\bar{\mathbf{a}})= -\frac{\frac{\partial^2 f}{\partial z^2}(\mathbf{a})\left(\frac{\partial f}{\partial x}(\mathbf{a})\right)^2 - 2 \frac{\partial^2 f}{\partial x\partial z}(\mathbf{a})\frac{\partial f}{\partial x}(\mathbf{a})\frac{\partial f}{\partial z}(\mathbf{a})+\frac{\partial^2 f}{\partial x^2}(\mathbf{a})\left(\frac{\partial f}{\partial z}(\mathbf{a})\right)^2}{\left(\frac{\partial f}{\partial z}(\mathbf{a})\right)^3},$$ as desired. $\square$