Differentiate both sides of the equation with respect to #x#. By doing this, a #dy/dx# should "appear" from the #y# variable in the equation. Isolating the #dy/dx# would give you the derivative implicitly.

Differentiating both sides with respect to #x#:

#D_x[-x^(2)y^(2)-3y^(3)+2]=D_x[5x^(3)]#

This will give you:

#[(-2x)(y^(2))+(2y*dy/dx)(-x^(2))]-9y^(2)*dy/dx+0=15x^(2)#

#-2xy^(2)-2yx^(2)*dy/dx-9y^(2)*dy/dx=15x^(2)#

Put all terms with #dy/dx# to the left and shove the other terms to the right.

#-2yx^(2)*dy/dx-9y^(2)*dy/dx=15x^(2)+2xy^(2)#

Factor #dy/dx# out.

#dy/dx*(-2yx^(2)-9y^(2))= 15x^(2)+2xy^(2)#

Divide everything by #(-2yx^(2)-9y^(2))#. You will get the derivative.

#dy/dx = (15x^(2)+2xy^(2))/(-2yx^(2)-9y^(2))#