The question tells us that the functions $x(t)$ and $y(t)$ satisfy $dx/dt=-x^2y$ and $dy/dt= -xy^2$ when $t=0$, $x=1$, and $y=2$. I have already worked out that $dy/dx= (xy^2)/(x^2y)$ and that $y=2x$. Now given that $dx/dt=-2x^3$ deduce a formula for $x$ as a function of $t$.
Any help would be much appreciated
You're almost there. You just have to separate the variables -
$$\frac{dx}{dt}=-2x^3$$ $$-\frac{dx}{2x^3}=dt$$ $$-\int\frac{dx}{2x^3}= t+C$$ $$\frac{1}{4x^2}=t+C$$ $$x=\pm\frac{1}{2\sqrt{t+C}}$$
$$1 = \pm\frac{1}{2\sqrt{C}}$$
Assuming $C\in\mathbb{R}$, this means $C=\frac{1}{4}$ and only the positive solution is admitted. Then $$x = \frac{1}{\sqrt{4t+1}}$$