Determining equation of the tangent to the surface with tricky powers I have a surface parametrised by
$$ x = u\sin^3v,\quad y=u\cos^3v,\quad z=u$$
and I'm trying to find the tangent plane at $(x_0, y_0, z_0) = (\sqrt2, -\sqrt2, 4)$
using this formula:
$$(x-x_0, y-y_0, z-z_0)\cdot \mathbf n(u_0,v_0)=0$$
I found the normal $ \mathbf n $ but am having difficulty finding $v_0$.
My attempt: 
$ x_0=u_0\sin^3v_0 $
$ \sin^3v_0 = \frac{\sqrt2}{4} $
Is this the correct method?

You know that:
\begin{align*}
 \frac{\sqrt{2}}{4} = \sin^3(v)\\
 \frac{-\sqrt{2}}{4} = \cos^3(v)
\end{align*}
Hence we know that $\tan(v)^3 = -1 \Leftrightarrow \tan(v)=-1$. The two solutions to this equation is $-\frac{\pi}{4}$ and $\pi - \frac{\pi}{4}$, where the latter is the solution that you are looking for. The rest you already explained how to do.
