Performing implicit differentiation involves differentiating each term on its own. We assume that #y# is a function of #x# so that any time we differentiate a function of #y#, we will still have our unknown #dy/dx# term left over.
#d/dx 2 = color(red)(d/dx (e^(xy^3 - x))) - color(green)(d/dx (y^2x^3)) + d/dx (y)#
The first term is a constant, therefore the derivative is #0#. The last term is simply #dy/dx#. For clarity, lets go over the other two terms individually.
For the #color(red)"red"# term, we can use the fact that #d/dx(e^f(x)) = e^f(x)f'(x)#.
#color(red)(d/dx (e^(xy^3 - x)) = (e^(xy^3 - x)) d/dx (xy^3 - x) )#
#color(red)(=(e^(xy^3 - x)) (d/dx(xy^3)-d/dx(x)))#
Apply the product rule, #d/dx f(x)g(x) = f'(x)g(x)+f(x)g'(x)#.
#color(red)(=(e^(xy^3 - x))(d/dx(x)*y^3 + x*d/dx(y^3) - 1))#
#color(red)(=(e^(xy^3 - x))(y^3 + 3xy^2 dy/dx - 1))#
The green term is easier, we just need to apply the product rule again.
#color(green)(d/dx(y^2x^3) = d/dx(y^2)*x^3 + y^2 d/dx(x^3))#
#color(green)(=2y dy/dx *x^3 + 3y^2x^2)#
Now we can put everything back together. Don't forget that the middle term was negative.
#0 = e^((xy^3 - x))(y^3 + 3xy^2 dy/dx - 1) - 2y dy/dx *x^3 - 3y^2x^2 +dy/dx#
Separate the terms that have #dy/dx# and move them to the left hand side.
#-e^((xy^3 - x))3xy^2 dy/dx+2x^3y dy/dx-dy/dx = e^((xy^3 - x))(y^3-1)- 3y^2x^2#
#(-e^((xy^3 - x))3xy^2+2x^3y -1)dy/dx = e^((xy^3 - x))(y^3-1)- 3y^2x^2#
Solve for #dy/dx#.
#dy/dx = -(e^((xy^3 - x))(y^3-1)- 3y^2x^2)/(e^((xy^3 - x))3xy^2+2x^3y -1)#
