Question: Please provide the steps to solve the equation, the answer to this equation is zero
I am not sure how it is derived, Kindly help

if $x+y=29$ then the value of $\dfrac{a}{x-a}+\dfrac{a}{y-a}$ is

Answer: 
Note that
$$
\frac{a}{x-a}+\frac{a}{y-a}=
\frac{a(y-a+x-a)}{(x-a)(y-a)}=\frac{a(x+y-2a)}{(x-a)(y-a)}
$$
which is zero if and only if either $a=0$ or $x+y=2a$.

So I guess you read wrongly your assignment when reporting it as $x+y=29$: look closely and you'll see it has $x+y=2a$.
I have got $$a\left(\frac{1}{x-a}+\frac{1}{29-x-a}\right)=a\left(\frac{29-x-a+x-a}{(x-a)(29-x-a)}\right)=\frac{a(29-2a)}{(x-a)(29-x-a)}$$
If you meantg that $$x+y=a$$ is given then we get $$\frac{a^2}{x(x-a)}$$