Let $G(x,y)=F(x,y,f(x,y))$. I want to find $\dfrac{\partial^2G}{\partial x^2}$ in terms of partial derivative of $F$ and $f$. I am not sure if my solution is correct as it looks different from the one in my textbook. The below is my solution: Put $z = f(x,y)$
\begin{align*} \dfrac{\partial^2 G}{\partial x^2} &= \dfrac{\partial}{\partial x}\left(\dfrac{\partial F}{\partial x} + \dfrac{\partial F}{\partial z} \dfrac{\partial z}{\partial x} \right) \\ &= \dfrac{\partial^2 F}{\partial x^2} + \dfrac{\partial F}{\partial z}\dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial f}{\partial x}\dfrac{\partial^2 F}{\partial x\partial z} \end{align*}
The solution in the textbook is: $$\dfrac{\partial^2 F}{\partial x^2} + \dfrac{\partial F}{\partial z}\dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial f}{\partial x}\left(\dfrac{\partial^2 F}{\partial x\partial z} + \dfrac{\partial^2 F}{\partial z\partial x} \right) + \left(\dfrac{\partial f}{\partial x}\right)^2\dfrac{\partial^2 F}{\partial z^2}$$
This notation makes a distinction between the differential operator ($\frac{\partial~~}{\partial x}$, which takes place with respect to $x$ where-ever it occurs in the composition to which it is being applied) and the partial differential expression (which is made with respect to $x$ as an argument of the function, while holding remaining arguments constrained (even if they too contain $x$)).
Thus the chain rule is implemented as:
$$\small\begin{align}\dfrac{\partial~~}{\partial x}F &= \dfrac{\partial F}{\partial x}+\dfrac{\partial f}{\partial x}\cdot\dfrac{\partial F}{\partial z}\\[1ex]\dfrac{\partial~~}{\partial x}\dfrac{\partial F}{\partial x}&=\dfrac{\partial^2 F}{\partial x~^2}+\dfrac{\partial f}{\partial x}\cdot\dfrac{\partial^2 F}{\partial z~\partial x}\\[1ex]\dfrac{\partial~~}{\partial x}\dfrac{\partial F}{\partial z}&=\dfrac{\partial^2 F}{\partial x~\partial z}+\dfrac{\partial f}{\partial x}\cdot\dfrac{\partial^2 F}{\partial z~^2}\end{align}$$
So...taking it slow
$$\small\begin{align}\dfrac{\partial^2 G}{\partial x~^2}&=\dfrac{\partial~~}{\partial x}\left(\dfrac{\partial~~}{\partial x}F\right)\\[1ex]&=\dfrac{\partial~~}{\partial x~}\left(\dfrac{\partial F}{\partial x}+\dfrac{\partial f}{\partial x}\cdot\dfrac{\partial F}{\partial z}\right)\\[1ex]&=\dfrac{\partial~~}{\partial x}\dfrac{\partial F}{\partial x}+\dfrac{\partial^2 f}{\partial x~^2}\cdot\dfrac{\partial F}{\partial z}+\dfrac{\partial f}{\partial x}\cdot\dfrac{\partial~~}{\partial x}\dfrac{\partial F}{\partial z}\tag 1\\[1ex]&=\left(\dfrac{\partial^2 F}{\partial x~^2}+\dfrac{\partial f}{\partial x}\cdot\dfrac{\partial^2 F}{\partial z~\partial x}\right)+\dfrac{\partial^2 f}{\partial x~^2}\cdot\dfrac{\partial F}{\partial z}+\dfrac{\partial f}{\partial x}\cdot\left(\dfrac{\partial^2 F}{\partial x~\partial z}+\dfrac{\partial f}{\partial x}\cdot\dfrac{\partial^2 F}{\partial z~^2}\right)\\[1ex]&=\dfrac{\partial^2 F}{\partial x~^2}+\dfrac{\partial f}{\partial x}\cdot\left(\dfrac{\partial^2 F}{\partial z~\partial x}+\dfrac{\partial^2 F}{\partial x~\partial z}\right)+\dfrac{\partial^2 f}{\partial x~^2}\cdot\dfrac{\partial F}{\partial z}+\left(\dfrac{\partial f}{\partial x}\right)^2\cdot\dfrac{\partial^2 F}{\partial z~^2}\tag 2\end{align}$$