An exercise on the implicit function theorem I am trying to learn the implicit function theorem and this is one exercise about it; I have solved it and would be grateful for any feedback on my solution, thanks.
Let $f\begin{pmatrix}x\\ y\\ z\end{pmatrix}=x y^2+\sin(xz)+e^z$ and $\textbf{a}=\begin{bmatrix}1\\ -1\\ 0 \end{bmatrix}$.
(a) Show that the equation $f=2$ defines $z$ as a $\mathcal{C}^1$ function $z=\phi\begin{pmatrix}x\\ y\end{pmatrix}$ near $\textbf{a}.$
(b) Find $\frac{\partial\phi}{\partial x}\begin{pmatrix}1\\-1\end{pmatrix}$ and $\frac{\partial\phi}{\partial y}\begin{pmatrix}1\\-1\end{pmatrix}.$
(c) Find the equation of the tangent plane of the surface $f^{-1}(\{2\})$ at $\mathbf{a}$ in two ways.
What I have done:
(a) Let $F=f-2$; $F$ is a $\mathcal{C}^1$ function, $F(\mathbf{a})=0,\ DF\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{bmatrix}y^2+z\cos(xz) &2xy & x\cos(xz)+e^z\end{bmatrix}$ and $DF(\mathbf{a})=DF\begin{pmatrix}1\\ -1\\ 0\end{pmatrix}=\begin{bmatrix}1 & -2 & 2\end{bmatrix}$ so in particular $\frac{\partial F}{\partial z}(\mathbf{a})=2\neq 0$ thus there exists a neighborhood V of $\begin{pmatrix}1\\-1\end{pmatrix}$ and $W$ of $0$ and a $\mathcal{C}^1$ function $\phi:V\to W$ so that $z\in W\Leftrightarrow z=\phi\begin{pmatrix}x\\ y\end{pmatrix},\ \begin{pmatrix}x\\ y\end{pmatrix}\in V.$
(b) $$\frac{\partial\phi}{\partial x}=-\frac{\frac{\partial F}{\partial x}\begin{pmatrix}1\\ -1\\ 0\end{pmatrix}}{\frac{\partial F}{\partial z}\begin{pmatrix}1\\ -1\\ 0\end{pmatrix}}=-\frac{1}{2}$$ and $$\frac{\partial\phi}{\partial y}=-\frac{\frac{\partial F}{\partial y}\begin{pmatrix}1\\ -1\\ 0\end{pmatrix}}{\frac{\partial F}{\partial z}\begin{pmatrix}1\\ -1\\ 0\end{pmatrix}}=-\frac{-2}{2}=1.$$
(c) Tangent plane: $$\begin{bmatrix}1 & -2 & 2\end{bmatrix} \begin{bmatrix}x-1\\ y+1\\ -z\end{bmatrix}=x-1-2(y+1)-2z=0\Leftrightarrow x-2y-2z=3$$

(a) and (b) are fine. As regards (c), there is a minor error in your work: the tangent plane should be
$$\begin{bmatrix}1 & -2 & 2\end{bmatrix} \begin{bmatrix}x-1\\ y+1\\ z-0\end{bmatrix}=x-1-2(y+1)+2z=0\Leftrightarrow x-2y+2z=3.$$
Here it is another way to find the tangent plane. Since $\phi$ is differentiable at $\begin{pmatrix}1\\-1\end{pmatrix}$, then
$$\phi\begin{pmatrix}x\\ y\end{pmatrix}=\phi\begin{pmatrix}1\\-1\end{pmatrix}+
\nabla \phi\begin{pmatrix}1\\-1\end{pmatrix}\cdot \left( \begin{pmatrix}x\\ y\end{pmatrix}-\begin{pmatrix}1\\-1\end{pmatrix}\right)+o\left(\sqrt{(x-1)^2+(y+1)^2}\right)$$
Hence the required tangent plane is
$$z= \phi\begin{pmatrix}1\\-1\end{pmatrix}+
\nabla \phi\begin{pmatrix}1\\-1\end{pmatrix}\cdot \left( \begin{pmatrix}x\\ y\end{pmatrix}-\begin{pmatrix}1\\-1\end{pmatrix}\right)=
0+
\begin{pmatrix}-\frac{1}{2}\\1\end{pmatrix}\cdot  \begin{pmatrix}x-1\\ y+1\end{pmatrix}\\
=-\frac{1}{2}(x-1)+(y+1)$$
which is equivalent to the equation given above.
