The parametric surface has the following parametric equation:
$$f(r,\theta)=(x,y,z)=(2r\cos\theta,5r\sin\theta,r).$$ Then we have
$$f_r(2,\pi/4)=(2\cos(\pi/4),5\sin(\pi/4),1)=(\sqrt{2},5\sqrt{2}/2,1),$$
$$f_\theta(2,\pi/4)=(-2\cdot 2\cdot\sin(\pi/4),5\cdot 2\cdot\cos(\pi/4),0)=(-2\sqrt{2},5\sqrt{2},0).$$
Therefore, the unit normal $n$ of the tangent plane at $(r,\theta)=(2,\pi/4)$ is equal to
$$n=\frac{f_r(2,\pi/4)\times f_\theta(2,\pi/4)}{\|f_r(2,\pi/4)\times f_\theta(2,\pi/4)\|}=\frac{1}{\sqrt{458}}(-5\sqrt{2},-2\sqrt{2},20).$$
Since the tangent plane at $(r,\theta)=(2,\pi/4)$ passes through the point $f(2,\pi/4)=(2\sqrt{2},5\sqrt{2},2)$, the equation of the tangent plane is given by
$n\cdot(x-2\sqrt{2},y-5\sqrt{2},z-2)=0$, or equivalently,
$$(-5\sqrt{2},-2\sqrt{2},20)\cdot(x-2\sqrt{2},y-5\sqrt{2},z-2)=0,$$
that is
$$5\sqrt{2}x+2\sqrt{2}y-20z=0.$$
This is a typical example of singular point(s) arising from the parametrization, they are called the artificial singularities.
For a parametrized surface: $S: \mathbf{r} = \mathbf{r}(u,v)$, like you said:
$$
\mathrm{If } \;\; \mathbf{r}_u \times \mathbf{r}_v \neq 0,
$$
then it indeed is a normal vector (un-normalized) to the surface $S$. Now:
$$
\frac {\partial P(\phi, \theta)}{\partial \phi} \times \frac {\partial P(\phi, \theta)}{\partial \theta} = \begin{vmatrix} \mathbf{i}& \mathbf{j} &\mathbf{k}
\\
\cos\phi \cos\theta & \cos\phi \sin\theta & -\sin\phi
\\
-\sin\phi \sin\theta & \sin\phi \cos\theta & 0 \end{vmatrix},
$$
notice the cross product is 0 for $\phi = 0,\pi$, this happens due to the choice of parametrization. For example, if you choose the angle between $y$-axis as the polar angle, then this singular point will be gone (other arises though), and this gives you a heuristics of one of the reasons why we wanna cut a manifold into pieces and establish a local coordinate system...
Best Answer
Regarding to what we know about the Curvilinear coordinates try to show that the Gradient vector can be described as follows in spherical coordinates: