$x^2 dy/dx + 2xy=y^{-3}$, find $y(x)$ The question states to start by substituting $u=y^{-2}$, however I am not sure how to deal with this as-as far as I can see there is no easy substitution. 
Are you meant to find that $y=(1/u)^{1/2}$ and then work with that(seems a little messy)? 
Thanks,

Rewrite it as follows
$$
(2xy-y^{-3})dx+x^2dy=0
$$
Multiply it by $x^6y^3$, then
$$
(2x^7y^4-x^6)dx+x^8y^3dy=0
$$
Which is full differential. Hence:
$$
F=\int{x^8y^3dy}=\frac{x^8y^4}{4}+f(x)\\
\frac{dF}{dx}=2x^7y^4+\frac{df}{dx}=2x^7y^4-x^6\to\\
f=-\frac{x^7}{7}\to F=\frac{x^8y^4}{4}-\frac{x^7}{7}=C^*
$$
Which represent family of soutions:
$$
y^4=\frac{C}{x^8}+\frac{4}{7x}
$$
Edit
Consider substitution $u=y^{-2}\to du=-2y^{-3}dy$, then we have following equation:
$$
-\frac{x^2y^3}{2}\frac{du}{dx}+2xy=y^{-3}
$$
Divide both sides of equation by $y^3$
$$
-\frac{x^2}{2}\frac{du}{dx}+2xy^{-2}=y^{-6}\to
-\frac{x^2}{2}\frac{du}{dx}+2xu=u^3\to\\
(u^3-2xu)dx+\frac{x^2}{2}du=0
$$
Multiply equation by $u^{-3}x^6$, then we have full differential
$$
(x^6-2x^7u^{-2})dx+\frac{x^8u^{-3}}{2}du=0
$$
From which 
$$
F=\int\frac{x^8u^{-3}}{2}du=-\frac{x^8u^{-2}}{4}+f(x)\\
\frac{dF}{dx}=-2x^7u^{-2}+\frac{df}{dx}=x^6-2x^7u^{-2}\to\\
f=\frac{x^7}{7}\to F=\frac{x^7}{7}-\frac{x^8u^{-2}}{4}=C^*
$$
Which after inverse substitution leads to the same family of solutions
$$
y^4=\frac{C}{x^8}+\frac{4}{7x}
$$
