The function $x^3 - 3x^2 - 6xy + 2y^2+ 8y + 8$ yields
$f_x = 3x^2 - 6x - 6y$
$f_y = 4y + 8 - 6x$
When we equate both of these to $0, y = \frac{x^2}{2} - x$ and $y = \frac{3}{2}x - 2$
Plotting these on a graph we can see the points of interaction are $(1, -0.5)$ and $(4, 4)$ and although that is the answer, I was wondering how to get those results using simultaneous equations, as the $x^2$ means that Gaussian elimination is inadequate, but I can't seem to figure it out using simultaneous equations.
Also, is the equation of a tangent plane the same formula as linear approximation? The tangent plane equation I found when at $(1, -2)$ is 13 + 9x - 6y
Just equate them. We need $y=\frac{x^2}2-x=\frac{3x}2-2$ giving $x^2-5x+4=0\iff x=1,4$ and corresponding $y=-1/2,4$.
$13+9x-6y$ is not an equation. For the functional equation $f(x,y,z)=0$, the equation of the tangent plane at $(x_0,y_0,z_0)$ is $[(x,y,z)-(x_0,y_0,z_0)]\cdot\vec\nabla f(x_0,y_0,z_0)=0$ where $\vec\nabla f=(f_x,f_y,f_z)$ represents the gradient or normal vector to the surface at a given point. Simplifying, you get$$(x-x_0)f_x(x_0,y_0,z_0)+(y-y_0)f_y(x_0,y_0,z_0)+(z-z_0)f_z(x_0,y_0,z_0)=0$$
We also have$$f(x-x_0,y-y_0,z-z_0)=\underbrace{\require{cancel}\cancelto{0}{f(x_0,y_0,z_0)}}_{(x_0,y_0,z_0)\text{ lies on the surface }f(x,y,z)=0}\\+(x-x_0)f_x(x_0,y_0,z_0)+(y-y_0)f_y(x_0,y_0,z_0)+(z-z_0)f_z(x_0,y_0,z_0)+\text{higher order terms}$$so indeed the equation of the tangent plane of $f(x,y,z)=0$ resembles the linear approximation of $f(x,y,z)$.