How to find the area inside $x^2-xy+y^2=1$? 
Find the area inside $x^2-xy+y^2=1$. Tip: $x^2-xy+y^2=\big(x-\frac{y}{2}\big)^2+\big(\frac{\sqrt 3y}{2}\big)^2$.

I thought change of variables would be appropriate here.
$$
u=x-\frac{y}{2}\\
v=\frac{\sqrt 3y}{2}
$$
Then we'd get $u^2+v^2=1$. This is a circle so its area is $\pi$. I guess we just need to multiply it by the Jacobian $J$ which is:
$$
\frac{1}{J}=\begin{bmatrix}
1&-0.5\\0&\sqrt 3/2
\end{bmatrix}=\sqrt 3/2\implies J=2/\sqrt 3
$$
Then the final answer would be $2/\sqrt 3\cdot \pi$.
Really not sure if this is correct.

The answer is correct. An easier approach is to realize the shape is an ellipse--the area of which is $\pi a b$. Setting $ x=y$ will lead to a $a  = \sqrt{2}$, setting $ x=-y$ will lead to $b  = \sqrt{2}/\sqrt{3}$, so $A = 2\pi/\sqrt{3}$.
