PDE Manipulation - Calculus I need help for this question, its a lot of calculus but I'm confuse. 
let $$ u= \dfrac{(x-b)^{2}+y^{2}-q^{2}}{(x-b-1)^{2}+y^{2}-q^{2}-1} $$
I need show that
$$ u_{x}^{2}+u_y^{2}= \dfrac{1}{(x-b)^{2}}\left(u-1 \right)^{2}\left( u^{2}+q^{2}(u-1)^{2}\right) $$
I begin with this
\begin{eqnarray*}u_x & =&  \dfrac{2(x-b)\left[ (x-b-1)^{2}+y^{2}-q^{2}-1\right]-2(x-b-1)\left[ (x-b)^{2}+y^{2}-q^{2}\right]}{\left[(x-b-1)^{2}+y^{2}-q^{2}-1\right]^{2}}\\\\
&=& \dfrac{2(x-b)A}{A^{2}} - \dfrac{2(x-b-1)B}{A^{2}}
\end{eqnarray*}
where $A = \left[(x-b)^{2}+y^{2}-q^{2}\right]$ and $B= \left[(x-b-1)^{2}+y^{2}-q^{2}-1\right]$
so 
\begin{eqnarray*}u_x & =&  \dfrac{2(x-b)}{A} - \dfrac{2(x-b-1)u}{A}.
\end{eqnarray*}
I do not know how can I do "appears q" and other manipulations, I've done the square that, I've done the same for the derivative with y, do not leave the place ...

It is a bit easier to write 
$$\begin{align}
u&=\frac{(x-b)^2+y^2-q^2}{(x-b-1)^2+y^2-q^2-1}&\\\\
&=\frac{(x-b-1)^2+y^2-q^2-1+2(x-b)}{(x-b-1)^2+y^2-q^2-1}\\\\&=1+2\frac{(x-b)}{(x-b-1)^2+y^2-q^2-1}\\\\
&=1+2\frac{s}{(s-1)^2+t^2}
\end{align}$$
where $s=x-b$ and $t^2=y^2-q^2-1$.
Using the chain rule, we have
$$\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 u}{\partial s^2}$$
and
$$\frac{\partial^2 u}{\partial y^2}=\frac{\partial^2 u}{\partial t^2}\left(\frac{dt}{dy}\right)^2+\frac{\partial u}{\partial t}\frac{d^2t}{dy^2}$$
Now, using $dt/dy=y/t$, $d^2t/dy^2=-(1+q^2)/t^3$ we find
$$\frac{\partial^2 u}{\partial y^2}=\frac{\partial^2 u}{\partial t^2}\left(\frac{t^2+(1+q^2)}{t^2}\right)-\frac{(1+q^2)}{t^3}\frac{\partial u}{\partial t}$$
The rest is just persevering.
