Matrices: $(X^{−1} + Y^{−1})^{−1} $ $(X^{−1} + Y^{−1})^{−1} = Y −Y(X + Y )^{−1}Y$
Here X, Y, and X + Y are nonsingular. 
The only I don't get is the left part, because you cannot get rid of the whole expression by taking $X + Y$ inverse because the X and Y in the expression are also inversed. But you can also not get rid of the x and y separate because they are in the $X+Y$ expression. 
Second thing aren't $X$ and $Y$ always nonsingular if $X+Y$ is nonsingular?
 A: First note that if you swap $X$ and $Y$ you should get the same inverse, that's because sum is a commutative operation, so both $[Y - Y(X + Y)^{-1}Y]$ and $[X -X(X + Y)^{-1}X]$ are inverse (and obviouly $[Y - Y(X + Y)^{-1}Y] = [X - X(X + Y)^{-1}X]$)
To prove it just multiply by your r.h.s expression
\begin{eqnarray}
(X^{-1} + Y^{-1}) (X^{-1} + Y^{-1})^{-1} &=& (X^{-1} + Y^{-1}) [Y - Y(X + Y)^{-1}Y] \\
&=&(X^{-1} + Y^{-1})Y[1 - (X+Y)^{-1}Y] \\
&=&X^{-1} X[1 - (X+Y)^{-1}X] + Y^{-1} Y[1 - (X+Y)^{-1}Y] \\
&=&1 - (X + Y)^{-1}X + 1 - (X+Y)^{-1}Y \\
&=& 1 - (X + Y)^{-1}(X + Y) + 1 \\
&=& 1
\end{eqnarray}
As for the second question, the answer is no. Imagine this case
$$
X = \pmatrix{1 & 0 \\ 0 & 0} ~~~ \mbox{and} ~~~ Y = \pmatrix{0 & 0 \\ 0 & 1}
$$
neither $X$ nor $Y$ are invertible, but $X + Y$ is
A: In the proof $$(X^{-1}+Y^{-1})^{-1}=((I+X^{-1}Y)Y^{-1})^{-1}=Y(I+X^{-1}Y)^{-1}=Y(I-(Y^{-1}(X+Y))^{-1})=Y(I-(X+Y)^{-1}Y),$$the third $=$ is the least trivial step, using$$(I+X^{-1}Y)(I-(Y^{-1}(X+Y))^{-1})=I+X^{-1}Y-X^{-1}Y(I+Y^{-1}X)(I+Y^{-1}X)^{-1}=I.$$This argument assumes $X,\,Y,\,X^{-1}+Y^{-1}$ are all invertible. (If the first two weren't, there'd be no such thing as $X^{-1}+Y^{-1}$ to invert in the first place.) All other requirements that a specific matrix be invertible follow from these. In particular, the inverse we seek to calculate is nonexistent iff some other inverse we need along the way is nonexistent.
