How to parametrize the surface $x^3 + 3xy + z^2 = 2$ and compute a tangent plane

Question

How do I parametrize the surface $x^3 + 3xy + z^2 = 2$ and compute the tangent plane at $(1, \frac{1}{3}, 0)$ using the resulting parametrization? I know that the tangent plane should be $$\nabla(x^3 + 3xy + z^2)\cdot(x - x_0, y - y_0, z - z_0) = 0 \implies \\ (3x_0^2 + 3y_0)(x - x_0) + 3x_0(y - y_0) + 2z_0(z - z_0) = 0,$$ which ends up being $4x + 3y = 5$ at $(1, \frac{1}{3}, 0)$. However, the parametrization I thought was right did not give me this answer (I'm not ruling out the possibility that I did the gradient method wrong).

Answer 1

Solve the equation for $y$. $$y=2/(3x) -(1/3)x^2 - (1/3)( z^2/x).$$

Let $u=x, z=v$. Then $$y=(2/3)(1/u)-(1/3)u^2-(1/3)(v^2/u).$$ Hence the surface may be parameterized by: $$F(u,v)=\left(u,\; (2/3)(1/u)-(1/3)u^2-(1/3)(v^2/u),\; v\right).$$ Compute tangent vectors of the parameterized surface $$Tu=\langle 1,\;(-2/3u^2)-(2u)/3+(v^2)/(3u^2),\; 0\rangle$$ for $u=x=1$ and $v=z=0$. Then $$Tu=<1, -4/3, 0>$$

$$Tv=<0, -2v/3u, 1>$$ for u=x=1 and v=z=0. Then $$Tv=<0, 0, 1>$$ Find the cross-product Tu x Tv =...= <-4/3, -1, 0> at (1, 1/3, 0) equation of the tangent line will be: (-4/3)(x-1)-1(x-1/3)+0(z-0)=0 ..... y=(-4/3)x+5/3