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Solution:

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, $\frac{\mathrm{du}}{\mathrm{dv}}=\frac{2}{3 \mathrm{x}}$