how can I find the tangent equation for $y=f(x)$ in $M(1,y_0)$ $(y_0 > 0)$?

Question

given function $y=f(x)$ so that $F(x,y)=0$, how can I find the tangent equation for $y=f(x)$ in $M(1,y_0)$ $(y_0 > 0)$?

$$2x^2-y^2+xy=0$$

I found that:

$$ y'=\frac{4+y_0}{2y_0 -1}$$

(I know that the answer is $y=2x$) and I don't know how to continue for solve it..

Answer 1

Note that

$$2x^2-y^2+xy=0\implies y^2-xy-2x^2=0\implies y=\frac{x\pm\sqrt{x^2+8x^2}}{2}\implies\begin{cases}y=2x\\y=-x\end{cases}$$