Let $f : U ⊆ \Bbb R^3 → R$ a differentiable function on $U$ an open of $\Bbb R^3$. If $a ∈ f_∗(U)$is a regular value of $f$, so $f^∗(a)$ is a regular surface in $\Bbb R^3$.
Try: Let $p=(x_0,y_0,z_0)\in f^*(a)$. Since $a$ is a regular value of $f$ and $a\in f_*(U)$, then $f_x(p)\ne 0,f_y(p)\ne 0$ or $f_z(p)\ne 0$. Wlog suppose that $f_z(p)\ne 0$. Define $F:U\subseteq \Bbb R^3\to \Bbb R^3$ by $F(x,y,z)=(x,y,f(x,y,z))$. We denote by $(u, v, t)$ the coordinates of a point of $\Bbb R^3$ where $F$ takes its values. Then, $$dF_p=\begin{bmatrix} 1 & 0 &0 \\ 0 & 1&0 \\ f_x&f_y &f_z \\ \end{bmatrix}$$ then $|dF_p|=f_z(p)\ne 0$.
By the inverse function theorem, there exist neighborhoods $V$ of $p$ and $W$ of $F(p)$ such that $F:V\to W$ is invertible and $F^{-1}:W \to V$ is differentiable. Then the component functions of $F^{-1}$, which are $x=u,y=v,z=g(u,v,t)$ are differentiable, since $F^{-1}$ makes it it is. Also the projection function of $V$ to the plane $xy$, $h(x,y)=g(u,v,t)=z$ is differentiable. Note that $$F(f^*(A)\cap V)=W\cap \{(u,v,t):t=a\}$$ I already tried this separately.
Now, notice that for $(u,v,t)$ $$(u,v,t)=(F\circ F^{-1})(u,v,t)=F(u, v,g(u,v,t))=(u,v,f(g(u,v,t))).$$ Therefore, $t=f(g(u,v,t))$ . What I need to prove is that $\text{graph}(h)=f^*(A)\cap V$, using the fact that $t=f(g(u,v,t))$ and $ F(f^*(a)\cap V)=W\cap \{(u,v,t):t=a\}$. I would need to test only that part to conclude my demonstration but I don't know how to do it. Someone help me.