Ridiculous 3 waysimultaneous equations need to get rid of $\lambda$ and find x and y $$x^2+8xy+7y^2-225=0
$$
$$2x+2\lambda x + 8\lambda y =0$$
$$2y+8\lambda x + 14 \lambda y = 0$$
So i need to equate for x and y by eliminating $\lambda$
i said $$x = \frac{-\lambda y}{2+2 \lambda }$$
$$y = \frac{-\lambda x}{2+14 \lambda }$$ 
From $f_x$ and $f_y$
 $$\lambda = \frac{-2x}{2x+8y}$$ 
 $$ \lambda = \frac{-2y}{8x+14y}$$ 
So solve...
$$ \frac{-2x}{2x+8y} = \ \frac{-2y}{8x+14y}$$
$$0=8y^2-8x^2-12yx$$
the remaining $xy$ really complicates things

$0=8y^2−8x^2−12yx$
$0 = -4(2x-y)(x+2y)$
so $x=-2y$ or $x=y/2$ would give the solution
