# x^2y^3 - 3y = x^2 #

Differentiating wrt #x# gives;

# d/dx(x^2y^3) - d/dx(3y) = d/dx(x^2) #

# :. d/dx(x^2y^3) - 3dy/dx = 2x #

We can use the product rule:

# d/dx(uv)=u(dv)/dx+v(du)/dx #

for the 1st term to give:

# (x^2)(d/dxy^3) + (y^3)(d/dxx^2) - 3dy/dx = 2x #

# :. (x^2)(d/dxy^3) + (y^3)(2x) - 3dy/dx = 2x #

And we use the chain rule so that we differentiate the 1st term wert #y# rather than #x# to give;

# :. (x^2)(dy/dxd/dyy^3) + 2xy^3 - 3dy/dx = 2x #

# :. (x^2)(dy/dx3y^2) + 2xy^3 - 3dy/dx = 2x #

# :. 3x^2y^2dy/dx + 2xy^3 - 3dy/dx = 2x #

# :. 3x^2y^2dy/dx - 3dy/dx = 2x - 2xy^3#

# :. 3(x^2y^2-1)dy/dx = 2x(1-y^3)#

# :. dy/dx = (2x(1-y^3))/(3(x^2y^2-1))#