Let $\alpha\in\mathbb{R}$ be fixed and let $f:\mathbb{R}^3\to\mathbb{R}$ be the (smooth) function $f(x,y,z)=(x^2+\alpha y^2)e^z-2\alpha$. If $\alpha>0$ note that the point $\bf{p}=(\alpha,\sqrt{\alpha},-\ln\alpha)$ lies in $S$. Find the equation of the tangent space $T_{\bf{p}}S$ in the form $ax+by+cz=0$.
Attempt: By definition, $T_{\bf{p}}S=\{c'(0)\in\mathbb{R}^3: c(t)\text{ a smooth curve in S with } c(0)=\bf{p}\}$. That is,\begin{equation}T_{{\bf{p}}}S=\{(x,y,z):\nabla f(\bf{p})\cdot (x,y,z)\}=0\end{equation}Now $\nabla f$$=(2xe^z,2\alpha ye^z,x^2+\alpha y^2)$, so we have $\nabla f(p)=(2,2\sqrt{\alpha}, 2\alpha^2)$.
Therefore $\nabla f(p)\cdot (x,y,z)=2x+2\sqrt{\alpha}y+2\alpha^2z=0$.
I am not fully confident this is correct - can anyone comment on this?
Hint:
$$\nabla f=(2xe^z,2\alpha ye^z,x^2+\alpha y^2\mathbf{e^z}),$$
so $\nabla f(p)=(2,2\sqrt{\alpha}, \mathbf{0})$, and $\nabla f(p)\cdot (x,y,z)=2x+2\sqrt{\alpha}y+0=0$.