what is the integral of dydx=y^2/x^2

[math_trouble]

im having problem integrating the equation

dydx=y^2/x^2

and also

dydx=3*y^2/x

[g_edgar]

I'm assuming they should be dy/dx=y^2/x^2
and dy/dx=3*y^2/x .

Your differential equations textbook should discuss "separation of variables" near the very beginning.

[StalkerM]

This can be rewritten in this way:
y'=y^2/x^2 with x different from zero.
y'/y^2=1/x^2
using chain rule:
d/dx[-1/y]=d/dx[-(1/x)+C]
consequentely:
1/y=(1/x)-C
y=1/[(1/x)+C]