Find the tangent plane to the image of $\phi(u,v)=(u^2,u\sin e^v,\frac{1}{3}u\cos e^v)$ at $(13,-2,1)$ It's clear we need to find the normal to the tangent vectors $T_u,T_v$, etc. But if we have to compute $T_u \times T_v$ at the specific point mentioned, we have to solve for $u,v$ in $(1)$:
$$\begin{cases} 13 = u^2 \\ -2 = u \sin e^v \\ 1 = \frac{1}{3} u \cos e^v \end{cases} \tag 1$$
Is this right? If there is no way around this, then how do we solve for $u,v$ in $(1)$?
 A: You don’t need to solve explicitly for both $u$ and $v$ (and in fact you don’t need to solve for either one). As you’ve found, $$T_u\times T_v = \left(-\frac13ue^v,\frac23u^2e^v\sin{e^v},2u^2e^v\cos{e^v}\right).$$ However, any nonzero scalar multiple of this vector will do for finding an equation of the tangent plane, so we pull out a factor of $ue^v$ (at the point we’re interested in $u\ne0$) and pull out a $1/3$ while we’re at it to make things more compact. So, the normal vector is now the simpler-looking $(-1,2u\sin{e^v},6u\cos{e^v})$. Setting $\phi(u,v)=(13,-2,1)$ gives you three simple equations in $u$, $\sin{e^v}$ and $\cos{e^v}$, which is exactly what you need to compute the scaled normal. Or, you might notice that $$(-1,2u\sin{e^v},6u\cos{e^v})=(-1,2y,18z),$$ so you don’t need to solve any equations at all!  
Another approach that’s a bit less work in this particular case is to derive an implicit Cartesian equation by eliminating $u$ and $v$. We have $y=u\sin{e^v}$ and $z=\frac13u\cos{e^v}$, so $y^2+9z^2=u^2=x$. Differentiating gives $\nabla\phi = (-1,2y,18z)$, which is what we ended up with above.
