Take one side at a time.
Let #f(x, y) = e^(x^2y)#.
#[(df)/(dx)]_"left" = (e^(x^2y))[x^2(dy)/(dx)+y*2x]#
(chain rule, product rule)
#[(df)/(dx)]_"right" = 2 + 2*(dy)/(dx)#
(chain rule with #(dy)/(dx)#)
Now, set them equal...
#(e^(x^2y))[x^2(dy)/(dx)+y*2x] = 2 + 2(dy)/(dx)#
...Multiply through.
#x^2e^(x^2y)(dy)/(dx) + 2xye^(x^2y) = 2 + 2(dy)/(dx)#
You can now isolate #(dy)/(dx)#. Move everything related to each other to each side.
#2xye^(x^2y) - 2 = 2(dy)/(dx) - x^2e^(x^2y)(dy)/(dx)#
Factor out common terms:
#2xye^(x^2y) - 2 = (dy)/(dx)[2 - x^2e^(x^2y)]#
Divide bracketed stuff over:
#[2xye^(x^2y) - 2]/[2 - x^2e^(x^2y)] = (dy)/(dx)#
...And simplify so it looks nicer.
#-[2 - 2xye^(x^2y)]/[2 - x^2e^(x^2y)] = (dy)/(dx)#
Wolfram Alpha lists the answer as:
#[2 - 2xye^(x^2y)]/[x^2e^(x^2y) - 2] = (dy)/(dx)#
