Answer
$\frac{∂w}{∂u}=\frac{2v^{2}[u^{2}v^{2}-(u-2v)^{2}]}{[u^{2}v^{2}+(u-2v)^{2}]^{2}}$
$\frac{∂w}{∂v}=\frac{u^{2}[(u-2v)^{2}-u^{2}v^{2}]}{[u^{2}v^{2}+(u-2v)^{2}]^{2}}$
Work Step by Step
Given that,
$w=\frac{rs}{r^{2}+s^{2}}; r=uv,s=u-2v$
Calculate the partial derivatives $\frac{∂w}{∂r},\frac{∂w}{∂s}$
$\frac{∂w}{∂r}=\frac{∂}{∂r}[\frac{rs}{r^{2}+s^{2}}]$ [Considering s as a constant]
$\frac{∂w}{∂r}=s.\frac{∂}{∂r}[\frac{r}{r^{2}+s^{2}}]$
$\frac{∂w}{∂r}=s.\frac{(r^{2}+s^{2})\frac{d}{dr}[r]-r\frac{∂}{∂r}[r^{2}+s^{2}]}{(r^{2}+s^{2})^{2}}$
$\frac{∂w}{∂r}=s.\frac{(r^{2}+s^{2}).1-r.(2r)}{(r^{2}+s^{2})^{2}}$
$\frac{∂w}{∂r}=s.\frac{(r^{2}+s^{2})-2r^{2}}{(r^{2}+s^{2})^{2}}$
$\frac{∂w}{∂r}=\frac{s.(s^{2}-r^{2})}{(r^{2}+s^{2})^{2}}$
$\frac{∂w}{∂s}=\frac{∂}{∂s}[\frac{rs}{r^{2}+s^{2}}]$ [Considering r as a constant]
$\frac{∂w}{∂s}=r.\frac{∂}{∂s}[\frac{s}{r^{2}+s^{2}}]$
$\frac{∂w}{∂s}=r.\frac{(r^{2}+s^{2})\frac{d}{ds}[s]-s\frac{∂}{∂s}[r^{2}+s^{2}]}{(r^{2}+s^{2})^{2}}$
$\frac{∂w}{∂s}=r.\frac{(r^{2}+s^{2}).1-s.(2s)}{(r^{2}+s^{2})^{2}}$
$\frac{∂w}{∂s}=r.\frac{(r^{2}+s^{2})-2s^{2}}{(r^{2}+s^{2})^{2}}$
$\frac{∂w}{∂s}=\frac{r.(r^{2}-s^{2})}{(r^{2}+s^{2})^{2}}$
Calculate the partial derivatives $\frac{∂r}{∂u},\frac{∂r}{∂v},\frac{∂s}{∂u},\frac{∂s}{∂v}$
$\frac{∂r}{∂u}=\frac{∂}{∂u}[uv]$ [Considering v as a constant]
$\frac{∂r}{∂u}=v$
$\frac{∂r}{∂v}=\frac{∂}{∂v}[uv]$ [Considering u as a constant]
$\frac{∂r}{∂v}=u$
$\frac{∂s}{∂u}=\frac{∂}{∂u}[u-2v]$ [Considering v as a constant]
$\frac{∂s}{∂u}=1$
$\frac{∂s}{∂v}=\frac{∂}{∂v}[u-2v]$ [Considering u as a constant]
$\frac{∂s}{∂v}=-2$
According to the chain rule for derivatives
$\frac{∂w}{∂u}=\frac{∂w}{∂r}\frac{∂r}{∂u}+\frac{∂w}{∂s}\frac{∂s}{∂u}$
$\frac{∂w}{∂u}=[\frac{s.(s^{2}-r^{2})}{(r^{2}+s^{2})^{2}}](v)+[\frac{r.(r^{2}-s^{2})}{(r^{2}+s^{2})^{2}}](1)$
$\frac{∂w}{∂u}=\frac{sv.(s^{2}-r^{2})}{(r^{2}+s^{2})^{2}}+\frac{r.(r^{2}-s^{2})}{(r^{2}+s^{2})^{2}}$
$\frac{∂w}{∂u}=\frac{sv.(s^{2}-r^{2})-r(s^{2}-r^{2})}{(r^{2}+s^{2})^{2}}$
$\frac{∂w}{∂u}=\frac{(s^{2}-r^{2})(sv-r)}{(r^{2}+s^{2})^{2}}$
Here $r=uv,s=u-2v$. So,
$\frac{∂w}{∂u}=\frac{[(u-2v)^{2}-(uv)^{2}][(u-2v)v-(uv)]}{[(uv)^{2}+(u-2v)^{2}]^{2}}$
$\frac{∂w}{∂u}=\frac{[(u-2v)^{2}-u^{2}v^{2}][uv-2v^{2}-uv]}{[u^{2}v^{2}+(u-2v)^{2}]^{2}}$
$\frac{∂w}{∂u}=\frac{(-2v^{2})[(u-2v)^{2}-u^{2}v^{2}]}{[u^{2}v^{2}+(u-2v)^{2}]^{2}}$
$\frac{∂w}{∂u}=\frac{2v^{2}[u^{2}v^{2}-(u-2v)^{2}]}{[u^{2}v^{2}+(u-2v)^{2}]^{2}}$
And,
$\frac{∂w}{∂v}=\frac{∂w}{∂r}\frac{∂r}{∂v}+\frac{∂w}{∂s}\frac{∂s}{∂v}$
$\frac{∂w}{∂v}=[\frac{s.(s^{2}-r^{2})}{(r^{2}+s^{2})^{2}}](u)+[\frac{r.(r^{2}-s^{2})}{(r^{2}+s^{2})^{2}}](-2)$
$\frac{∂w}{∂v}=[\frac{su.(s^{2}-r^{2})}{(r^{2}+s^{2})^{2}}]-[\frac{2r.(r^{2}-s^{2})}{(r^{2}+s^{2})^{2}}]$
$\frac{∂w}{∂v}=\frac{su(s^{2}-r^{2})-2r(r^{2}-s^{2})}{(r^{2}+s^{2})^{2}}$
$\frac{∂w}{∂v}=\frac{su(s^{2}-r^{2})+2r(s^{2}-r^{2})}{(r^{2}+s^{2})^{2}}$
$\frac{∂w}{∂v}=\frac{(s^{2}-r^{2})(su+2r)}{(r^{2}+s^{2})^{2}}$
Here $r=uv,s=u-2v$. So,
$\frac{∂w}{∂v}=\frac{[(u-2v)^{2}-(uv)^{2}][(u-2v)u+2(uv)]}{[(uv)^{2}+(u-2v)^{2}]^{2}}$
$\frac{∂w}{∂v}=\frac{[(u-2v)^{2}-u^{2}v^{2}][u^{2}-2uv+2uv]}{[u^{2}v^{2}+(u-2v)^{2}]^{2}}$
$\frac{∂w}{∂v}=\frac{u^{2}[(u-2v)^{2}-u^{2}v^{2}]}{[u^{2}v^{2}+(u-2v)^{2}]^{2}}$
