A simple differential equation

$\frac{dy}{dx} = x y^2$ $\frac{dy}{y^2} = xdx$ $\int_{f(1)}^{-1/3} \frac{dy}{y^2} = \int_{1}^{2} x dx$ $- \bigg[ \frac{1}{y} \bigg]_{f(1)}^{-1/3} = \frac{1}{2} \bigg[ x^2 \bigg]_{1}^{2}$ $- \bigg( \frac{1}{-1/3} - \frac{1}{f(1)} \bigg) = \frac{1}{2} \bigg( 2^2 - 1^2 \bigg)$ $f(1) = - \frac{2}{3}$

Let $y = f(x)$ . Then we have:

$\begin{aligned} f'(x) & = x\left(f(x)\right)^2 \\ \frac {dy}{dx} & = x y^2 \\ \frac 1{y^2} dy & = x \ dx \\ \int \frac 1{y^2} dy & = \int x \ dx \\ - \frac 1y & = \frac {x^2}2 + \color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ - \frac 1{-\frac 13} & = \frac {2^2}2 + C & \small \color{#3D99F6} \text{Since } f(2) = - \frac 13, \implies C = 1 \\ - \frac 1{f(x)} & = \frac {x^2}2 + 1 \\ \implies - \frac 1{f(1)} & = \frac 12 + 1 \\ f(1) & = \boxed{-\dfrac 23} \end{aligned}$

Start off by dividing both sides by $f(x)^2$ to get:

$\frac{f'(x)}{[f(x)]^2}=x$

$\Rightarrow$ $-\frac{1}{f(x)}=\frac{x^2}{2}+c$ $(1)$ (integrating both sides w.r.t $x$ )

Since $f(2)=-\frac{1}{3}$ , we can plug this into $(1)$ to find c

$\Rightarrow c=1$ $\Rightarrow -\frac{1}{f(x)} = \frac{x^2}{2} + 1$ $(2)$

Plugging $x=1$ into $(2)$ and after a little bit of simple algebra should give $f(1)=-\frac{2}{3}$

Done! A bite-sized problem to start off the day, I guess