What is the general solution of the differential equation # dy/dx=(y^3 - yx^2) / (x^3 + y^2x) #?

Presumably you require the solution, of the Differential equation rather than the equation, which is of course in the question.

We have:

# dy/dx=(y^3 - yx^2) / (x^3 + y^2x) # ..... [A]

Note, that we cannot isolate terms in #x# and #y# alone so let us attempt some manipulation in an attempt to simplify the equation:

# dy/dx = (y(y^2 - x^2)) / (x(y^2 + x^2)) #
# \ \ \ \ \ = (y/x) (y^2 - x^2) / (y^2 + x^2) xx (1/x^2)/(1/x^2)#
# \ \ \ \ \ = (y/x) ((y/x)^2 - 1) / ((y/x)^2+1) # ..... [B]

So Let us try a substitution, Let:

# v = y/x => y=vx => dy/dx = v + x(dv)/dx #

And substituting into the above DE [B], to eliminate #y#:

# v + x(dv)/dx = (v) (v^2 - 1) / (v^2+1) #

# :. x(dv)/dx = v{ (v^2 - 1) / (v^2+1) - 1 }#

# :. x(dv)/dx = v{ ( (v^2-1) - (v^2+1) ) / (v^2+1) }#

# :. x(dv)/dx = ( -2v ) / (v^2+1)#

# :. (v^2+1)/v (dv)/dx = -2/x #

Which has indeed helped as it has reduced the initial equation [A] to a separable equation, so we can now "seperate the variables" to get:

# int \ (v^2+1)/v \ dv = int \ -2/x \ dx#

# :. int \ v+1/v \ dv = int \ -2/x \ dx#

And this is trivial to integrate to get:

# 1/2v^2 + ln |v| = -2ln|x| - C #

And restoring the substitution we get:

# 1/2(y/x)^2 + ln |y/x| = -2ln|x| - C #
# :. y^2/(2x^2) + ln |y| -ln|x| = -2ln|x| - C #
# :. y^2/(2x^2) + ln |y| +ln|x| + C = 0#
# :. y^2/(2x^2) + ln |xy| + C = 0# QED