Let $f:\mathbb{R}^{2}\to \mathbb{R}$ be a twice continuously differentiable function such that $f(0,0)=0$ and $f_{y}(0,0)\ne 0$.
By the implicit function theorem, there exists an $\epsilon >0$ and a continuously differentiable function $\psi:(-\epsilon,\epsilon)\to \mathbb{R}$ such that $\psi(0)=0$ and $f(x,\psi(x))=0$ for all $x \in (-\epsilon,\epsilon)$.
Using the function $\psi$ constructed above, prove that the mapping $$F(x,w)=\left(x+wf_{x}(x,\psi(x)),\psi(x)+wf_{y}(x,\psi(x) \right)$$ is one-to-one in a neighbourhood of the origin $(0,0) \in \mathbb{R}^{2}$.
I tried to use the inverse function theorem. If I can show that $\det JF|_{(0,0)}\ne 0$, then the result will follow. The inverse function theorem says that, assuming $F$ is continuously differentiable in a neighbourhood of the origin, if $\det JF \ne 0$ at the origin, then $F$ has a smooth inverse in some open neighbourhood of the origin, so in particular, it is one-to-one.$$\det JF(x,w)=\begin{vmatrix} 1+w \cdot \left(f_{xx}(x,\psi(x))+f_{xy}(x,\psi(x))\frac{d}{dx}\psi(x) \right) & f_{x}(x,\psi(x)) \\ \frac{d}{dx}\psi(x)+w \cdot \left(f_{yx}(x,\psi(x))+f_{yy}(x,\psi(x))\frac{d}{dx}\psi(x)\right)& f_{y}(x,\psi(x)) \end{vmatrix}$$Moreover there is a hint in the question, which is to use $\frac{d}{dx}\psi(x)=-\frac{f_x(x,\psi(x))}{f_y(x,\psi(x))}$.
You have computed $\det JF$ for general $(x, w)$. To apply the inverse function theorem in the way you describe, you just need to evaluate $\det JF$ at the specific point $(x, w) = (0, 0)$, and check that the answer is non-zero.
If you do this, you find that \begin{align} \det JF(0,0) &=\begin{vmatrix} 1 & f_{x}(0,0)\\ \frac{d\psi}{dx}(0)& f_{y}(0,0) \end{vmatrix} \\ & = f_{y}(0,0) - \frac{d\psi}{dx}(0) \times f_{x}(0,0) \\ &= f_{y}(0,0) + \frac{f_{x}(0,0)^2}{f_y(0,0)} \\ &= \frac{f_y(0,0)^2 + f_x(0,0)^2 }{f_y(0,0)}\\ & \neq 0. \end{align}
(In this calculation, I used the facts that $\psi(0) = 0$ and $f_y(0, 0) \neq 0$.)