Answer
${\frac {\partial u}{\partial x}}=0$.
${\frac {\partial u}{\partial y}}=\frac{z}{(z-y)^2}$.
${\frac {\partial u}{\partial z}}=\frac{-y}{(z-y)^2}$
${\frac {\partial u}{\partial x}}$($\sqrt{3}$ ; $2$ ; $1)=0$.
${\frac {\partial u}{\partial y}}$($\sqrt{3}$ ; $2$ ; $1)=\frac{1}{(1-2)^2}=1$.
${\frac {\partial u}{\partial z}}$($\sqrt{3}$ ; $2$ ; $1)=\frac{-2}{(1-2)^2}=-2$
Work Step by Step
a) We express ${\frac {\partial u}{\partial x}}$,${\frac {\partial u}{\partial y}}$, and ${\frac {\partial u}{\partial z}}$
1) We apply functions of $x$,$y$ and $z$ by using the Chain Rule:
${\frac {\partial u}{\partial x}}$ = ${\frac {\partial u}{\partial p}}$$\frac{dp}{dx}$ + ${\frac {\partial u}{\partial q}}$$\frac{dq}{dx}$ + ${\frac {\partial u}{\partial r}}$$\frac{dr}{dx}$ = ${\frac {\partial }{\partial p}}$$(\frac{p-q}{q-r})$ $\frac{d}{dx}$$(x+y+z)$+${\frac {\partial }{\partial q}}$$(\frac{p-q}{q-r})$ $\frac{d}{dx}$$(x-y+z)$+${\frac {\partial }{\partial r}}$$(\frac{p-q}{q-r})$ $\frac{d}{dx}$$(x+y-z)$ = $\frac{1}{q-r}$ + $\frac{r-p}{(q-r)^2}$+$\frac{p-q}{(q-r)^2}$=$\frac{0}{(q-r)^2}$=0.
${\frac {\partial u}{\partial y}}$ = ${\frac {\partial u}{\partial p}}$$\frac{dp}{dy}$ + ${\frac {\partial u}{\partial q}}$$\frac{dq}{dy}$ + ${\frac {\partial u}{\partial r}}$$\frac{dr}{dy}$ = ${\frac {\partial }{\partial p}}$$(\frac{p-q}{q-r})$ $\frac{d}{dy}$$(x+y+z)$+${\frac {\partial }{\partial q}}$$(\frac{p-q}{q-r})$ $\frac{d}{dy}$$(x-y+z)$+${\frac {\partial }{\partial r}}$$(\frac{p-q}{q-r})$ $\frac{d}{dy}$$(x+y-z)$ = $\frac{1}{q-r}$ - $\frac{r-p}{(q-r)^2}$+$\frac{p-q}{(q-r)^2}$=$\frac{2(p-r)}{(q-r)^2}$
where $p=x+y+z$, $q=x-y+z$ and $r=x+y-z$
${\frac {\partial u}{\partial y}}$=$\frac{2(x+y+z-x+y-z)}{(x-y+z-x-y+z)^2}$=$\frac{4z}{4(z-y)^2}$=$\frac{z}{(z-y)^2}$.
${\frac {\partial u}{\partial z}}$ = ${\frac {\partial u}{\partial p}}$$\frac{dp}{dz}$ + ${\frac {\partial u}{\partial q}}$$\frac{dq}{dz}$ + ${\frac {\partial u}{\partial r}}$$\frac{dr}{dz}$ = ${\frac {\partial }{\partial p}}$$(\frac{p-q}{q-r})$ $\frac{d}{dz}$$(x+y+z)$+${\frac {\partial }{\partial q}}$$(\frac{p-q}{q-r})$ $\frac{d}{dz}$$(x-y+z)$+${\frac {\partial }{\partial r}}$$(\frac{p-q}{q-r})$ $\frac{d}{dz}$$(x+y-z)$ = $\frac{1}{q-r}$ + $\frac{r-p}{(q-r)^2}$-$\frac{p-q}{(q-r)^2}$=$\frac{2(q-p)}{(q-r)^2}$
where $p=x+y+z$, $q=x-y+z$ and $r=x+y-z$
${\frac {\partial u}{\partial y}}$=$\frac{2(x-y+z-x-y+z)}{(x-y+z-x-y+z)^2}$=$\frac{-4y}{4(z-y)^2}$=-$\frac{y}{(z-y)^2}$.
2) We express $u$ directly in terms of $x$,$y$, and $z$ before differentiating:
u=$\frac{p-q}{q-r}$
where
$p=x+y+z$, $q=x-y+z$ and $r=x+y-z$
u=$\frac{x+y+z-x+y-z}{x-y+z-x-y+z}$=$\frac{y}{z-y}$.
${\frac {\partial u}{\partial x}}$=${\frac {\partial }{\partial x}}$$(\frac{y}{z-y})$=0.
${\frac {\partial u}{\partial y}}$=${\frac {\partial }{\partial y}}$$(\frac{y}{z-y})$=$\frac{{\frac {\partial }{\partial y}}[y](z-y)-y{\frac {\partial }{\partial y}[z-y]}}{(z-y)^2}$=$\frac{z-y+y}{(z-y)^2}$=$\frac{z}{(z-y)^2}$.
${\frac {\partial u}{\partial z}}$=${\frac {\partial }{\partial z}}$$(\frac{y}{z-y})$=-y$\frac{{\frac {\partial }{\partial z}}[z-y]}{(z-y)^2}$=-y$\frac{1}{(z-y)^2}$=$\frac{-y}{(z-y)^2}$.
b) We evaluate ${\frac {\partial u}{\partial x}}$,${\frac {\partial u}{\partial y}}$, and ${\frac {\partial u}{\partial z}}$ at the given point ($x$,$y$,$z$):
($x$,$y$,$z$)=($\sqrt{3}$ ; $2$ ; $1$)
Thus:
${\frac {\partial u}{\partial x}}$($\sqrt{3}$ ; $2$ ; $1$)=0.
${\frac {\partial u}{\partial y}}$($\sqrt{3}$ ; $2$ ; $1$)=$\frac{1}{(1-2)^2}$=1.
${\frac {\partial u}{\partial z}}$($\sqrt{3}$ ; $2$ ; $1$)=$\frac{-2}{(1-2)^2}$=-2.
