The function $f(x,y)=xy$ has an absolute maximum value and an absolute minimum value subject to the constraint $x^2+y^2-xy=9$.
I know $grad(f(x,y))=(y,x)$ and $grad(g(x,y) = (2x-y,2y-x)$
So how do I find these? I got about as far as $y = -2xλ^2 + 4yλ^2-yλ$ by substituting the x equation into the y equation but I have no idea what to do next because the explanations I've been able to find are gibberish or skipping steps.
You have the system$$\left\{\begin{array}{l}y=\lambda(2x-y)\\x=\lambda(2y-x)\\x^2+y^2-xy=9.\tag1\end{array}\right.$$The first two equations are equivalent to a system of two linear homogeneous equations in two unknowns, depending on the parameter $\lambda$:$$\left\{\begin{array}{l}-2\lambda x+(1+\lambda)y=0\\(1+\lambda)x-2\lambda y=0.\end{array}\right.$$If the determinant of the matrix of coefficients of this system is not $0$, then its only solution is $x=y=0$, but $(0,0)$ is not a solution of $x^2+y^2-xy=9$. On the other hand, the determinant of the matrix of coefficients of this system is $3\lambda^2-2\lambda-1$, whose roots are $1$ and $-\frac13$. So, all you have to do is to solve the system $(1)$ in these two cases. When $\lambda=1$, you shall get $(x,y)=\pm(3,3)$, and when $\lambda=-\frac13$, you shall get $(x,y)=\pm\left(\sqrt3,-\sqrt3\right)$.