Find the tangent plane to $$z=4 x^3+3 xy +4 y^3$$ at $(-1,1,-3)$.
Answer on the form $z=Ax+By+C$.
I don't know how to solve these problems, should I find the derivative of x and y, then let those be the vectors on my plane?
If I see how this problem is solved, then I might be able to solve similar ones.
$$\nabla z=\left\langle\frac{\partial z}{\partial x},\frac{\partial z}{\partial y}\right\rangle^T=\langle12x^2+3y,12y^2+3x\rangle^T$$ At $(-1,1,-3)$, we have $\nabla z=\begin{pmatrix}15\\9\end{pmatrix}$ so we can conclude that the plane is of the form $z=15x+9y+c$. Solving for $c$, we have $$\begin{align}-3&=15(-1)+9(1)+c\\3&=c\end{align}$$ So the tangent plane is $z=15x+9y+3$.
Your surface is defined by a polynomial equation $f(x,y,z) = 0$. You could interpret this situation geometrically in the following way:
$f:R^3\to R$ gives a color to every point of the 3D space (imagine you have a color palette indexed by real numbers). Assume white is indexed at $0$, then your surface is the locus of points with white color.
The gradient $$ \nabla f = ( \partial_x f, \partial_y f, \partial_z f ) $$ is orthogonal to the surface when evaluated at points $(x,y,z)$ of the surface. This is because the color stays constant in our white surface.
So it's enough for you to compute $\nabla f$ with partial derivatives, evaluate it at your point $(-1,1,-3)$ to get a vector $V$, and then notice that the set of vectors perpendicular to that one is given by the equation $$ V\cdot (X,Y,Z) = 0 $$ Doing this you find the equation of a plane parallel to the tangent space, but passing through the origin. To get the tangent space, translate it via $(-1,1,-3)$. Finally, you should get $$ Z=15X+9Y+3 $$
