Question

Find an equation of the tangent line to the graph at the given point. (4,-2(square rt of 3)
x^2y^2 -9x^2-4y^2 = 0,
This is what i did to find dy/dx
2x(y^2)-x^2(2dy/dx)-18x-8ydy/dx=0
-x^2(2ydy/dx)-8ydy/dx=-2xy^2-18x
(dy/dx)(-x^2 2y-8y)=-2xy^2-18x
dy/dx=(-2xy^2-18x)/(-2x^2y-8y)
i was wondering if i messed up the dy/dx somewhere