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I am given $x=u+v$, $y=3u^2$, and $z=u-v$. I need to find the equation of the tangent plane at $(2,3,0)$. I understand that the equation of the tangent plane is $z=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$, since $z_0=0$. I have determined that $u=1$ and $v=1$ at $(2,3,0)$.

I'm confused as to how I determine $f_x$ and $f_y$ since my coordinates are in terms of $u$ and $v$.

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The problem here is that your formula is for the tangent plane to a graph of the form $z = f(x, y)$, but your surface is expressed parametrically as $$ S: \mathbb R \times \mathbb R \mapsto \mathbb R^3: (u, v) \mapsto (u+v, 3u^2, u-v) $$

For that, the equation looks more like

$$ \begin{bmatrix} x - x_0 \\ x - y_0 \\ z - z_0 \end{bmatrix} \cdot \mathbf w = 0 $$ where $$ \mathbf w = \frac{\partial f}{\partial u} (u_0, v_0) \times \frac{\partial f}{\partial v} (u_0, v_0). $$

The two partials are tangent to your surface, so their cross product is normal to the tangent plane.

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