The slope of the tangent to the curve $10x^3+5x^2y+4xy^2+6y^3=25$ I am having problems understanding how to solve this equation: 
The slope of the tangent to the curve:
 $$10x^3+5x^2y+4xy^2+6y^3=25$$
at the point $(1,1)$.
Any help would be appreciated, thanks!

Hint: $$\dfrac{dy}{dx}=-\dfrac{F'_{x}}{F'_{y}}=\dfrac{30x^2+10xy+4y^2}{5x^2+8xy+18y^2}$$
so
$$y'_{x}|_{x=1}=-\dfrac{44}{31}$$
then the tangent
$$y-1=-\dfrac{44}{31}(x-1)$$
where
$$F=10x^3+5x^2y+4xy^2+6y^3-25$$
