Find $\displaystyle \frac{\partial y}{\partial x}$ at the point (1,-1) for $\displaystyle 3xy^{2}+5x^{2}y+4x=0$

So I came up with:

$\displaystyle (-3y^{2}\frac{\partial x}{\partial x}-6xy\frac{\partial y}{\partial x})+(10xy\frac{\partial x}{\partial x}+5x^{2}\frac{\partial y}{\partial x}+4\frac{\partial x}{\partial x})=0$

After cleaning it up a bit, I got:

$\displaystyle \frac{\partial y}{\partial x}(-6xy+5x^{2})=3 y^{2}-10xy-4$

$\displaystyle =>\frac{\partial y}{\partial x}=\frac{(3 y^{2}-10xy-4)}{(-6xy+5x^{2})}$

So:

$\displaystyle \frac{(3 y^{2}-10xy-4)}{(-6xy+5x^{2})}|_{(1,-1)}$

$\displaystyle =>\frac{(3 (-1)^{2}-10(1)(-1)-4)}{(-6(1)(-1)+5(1)^{2})}=\frac{9}{11}$

Can someone please double check this for me? Thanks