$z = 4(x-1)^2 + 5(y+3)^2 +1$ at the point $(2,-2,10)$
I'm not sure how exactly how to proceed through the problem. I know to find the derivative with respect to $x$ , $y$, and $z$, which I did. For $x$, I got $8(x-1)$ and for $y$, I got $10(y+3)$. When I found $z$, I got was pretty much the sum of the derivatives of $x$ and $y$ . I know I'm supposed to plug in the point $(2,-2,10)$ somewhere, but I'm not sure if it is the original question or the derived formulas.
Recall that the gradient is normal to the level surface. Therefore,
$$\begin{align} \vec N&=\nabla f(\vec r)\\\\ &=\hat x\frac{\partial f(x,y)}{\partial x}+\hat y\frac{\partial f(x,y)}{\partial y}\\\\ &=\hat x(8(x-1))+\hat y(10(y+3)) \end{align}$$
At the point $(2,-2,10)$, the normal is
$N=\hat x8+\hat y10$. The equation of the tangent plane is then
$$\vec N\cdot (\vec r-\vec r_0)\implies 4x+5y=-2$$