dy/dx = xy^2/x+5
how do I algebraically convert this into the form y/x
I'm not really sure what you are asking here. You aren't going to be able to express this as f(y/x).
The method you would use to solve this is called separation of variables.
$\displaystyle \frac{\text{dy}}{\text{dx}}=\frac{x y^2}{x+5}$
$\displaystyle \frac{\text{dy}}{y^2}=\frac{\text{dx} x}{x+5}$
And you integrate on both sides to obtain your solution. You should be able to do this. Post back if you need more help.
It's not hard to check it. The derivative of the left side, with respect to x, is y^{-2} dy/dx and the derivative of the right side is 1/(x+ 5).
So y^{-2} dy/dx= 1/(x+ 5) so dy/dx= y^2/(x+ 5).
I do NOT see an "x" in the numerator on the right.
The integral of $\displaystyle x/(x+ 5)$ is NOT ln(x+ 5). If you need to, make the substitution u= x+ 5 so that x= u- 5 and $\displaystyle \frac{x}{x+ 5}= \frac{u- 5}{u}= 1- \frac{1}{u}$.