Find a parametrization of the surface $$\ x^3 +3xy + z^2 =2; z0$$
and use it to find tangent plane at the point $$\ x = 1, y=\frac{1}{3}, z = 0 $$.

Now, I used $$\ u = x; v = y$$ so $$\ z=\sqrt{2-u^3-3uv} $$
and i got the normal to the surface: $$\ ( \frac{3(u^2 +v)}{2 \sqrt{2-u^3 - 3uv}} ,\frac{3u}{2\sqrt{2-u^3 -3uv}} , 1) $$
So, at the point u = 1and v = 1/3 my normal vector goes to $$(\infty,\infty,1)$$.

So how to deal with that?

Question

Find a parametrization of the surface $$\ x^3 +3xy + z^2 =2; z>0$$
and use it to find tangent plane at the point $$\ x = 1, y=\frac{1}{3}, z = 0 $$.

Now, I used $$\ u = x; v = y$$ so $$\ z=\sqrt{2-u^3-3uv} $$
and i got the normal to the surface: $$\ ( \frac{3(u^2 +v)}{2 \sqrt{2-u^3 - 3uv}} ,\frac{3u}{2\sqrt{2-u^3 -3uv}} , 1) $$
So, at the point u = 1and v = 1/3 my normal vector goes to $$(\infty,\infty,1)$$.

So how to deal with that?