We are given
$$y_1'=-\frac{y_1^{\frac{1}{3}}y_2}{x^2}$$ $$y_2'=y_1x^2$$
I tried at first differentiating the first equation wrt $x$ to give me $y_1''$ in terms of $y_1$, $y_1'$ and $y_2$ (which we can write using the 1st equation in terms of $y_1$ and $y_1'$). After some rearranging I am left with:
$$3xy_1'' + \frac{xy_1'^2}{y_1} + (6-3x)y_1'=0$$
Here is where I became rather mystified. Dividing through by $x$ doesn't seem to enlighten me, and I can't spot any total derivatives to bunch up the right 2 terms together. Any pointers?