Differentiate $x^{2}$ with respect to $x^{3}$.
Let $u=x^{2}$ and $v=x^{3}$
We need to differentiate $u$ with respect to $v$ that is find $\frac{d u}{d v}$.
On differentiating $u$ with respect to $x$, we get
$\frac{d u}{d x}=\frac{d}{d x}\left(x^{2}\right)$
We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1}$
$\Rightarrow \frac{d u}{d x}=2 x^{2-1}$
$\therefore \frac{\mathrm{du}}{\mathrm{dx}}=2 \mathrm{x}$
Now, on differentiating $v$ with respect to $x$, we get
$\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{3}\right)$
$\Rightarrow \frac{d v}{d x}=3 x^{3-1}$ (using the same formula)
$\therefore \frac{\mathrm{dv}}{\mathrm{dx}}=3 \mathrm{x}^{2}$
We have $\frac{\mathrm{du}}{\mathrm{dv}}=\frac{\frac{\mathrm{du}}{\mathrm{dx}}}{\frac{\mathrm{dv}}{\mathrm{dx}}}$
$\Rightarrow \frac{\mathrm{du}}{\mathrm{dv}}=\frac{2 \mathrm{x}}{3 \mathrm{x}^{2}}$
$\therefore \frac{\mathrm{du}}{\mathrm{dv}}=\frac{2}{3 \mathrm{x}}$
Thus, $\frac{\mathrm{du}}{\mathrm{dv}}=\frac{2}{3 \mathrm{x}}$