Find a parametrization of the surface $x^3 + 3xy + z^2 = 2$, $z > 0$, and use it to find the tangent plane at $x = 1$, $y = \dfrac{1}{3}$, $z = 0$.
I know how to find the tangent plane once I have the parametrization - it's the first part that's troubling me. I started by solving for $z$, which gave me the parametrization $(u, \, v, \, \sqrt{2-u^3-3uv})$. Then the partial derivative w.r.t. $u$ is $\left( 1, \, 0, \, \dfrac{-3(u^2+v)}{2 \sqrt{2-u^3 - 3uv}} \right)$. But when I plug in $u = 1$ and $y = \dfrac{1}{3}$, I get a discontinuity.
Not really sure what else to try. Maybe cylindrical coordinates would work better?
It is much easier to work without a parametrization. The gradient of a implicitly given surface is the normal of the tangent plane, no need for parametrization.
The gradient becomes
$$(3x^2+3y,3x,2z)=(4,3,0)$$
which gives you the normal.
Your parametrization is bad, because there are two branches of the square root and you are located precisely at $z=0$, where the branches meet. If you really want to use parametrization, select a different pair of variables. Or, you can multiply your entire divergent derivative vector by the square root in the denominator. Scaling doesn't matter if you only need directions.