Let $$f(x)=\frac{x}{(2x-1)}$$ and $$g(x)=\frac{x}{(x-1)}$$. Then the domain of the function $$h(x)=f(g(x))+g(f(x))$$ is all real numbers except

We check where $$f(x)=\frac{x}{2x-1}$$ or $$g(x)=\frac{x}{x-1}$$, or their compositions, become undefined.

First, f(x) is undefined at $$x=\tfrac12$$ and g(x) is undefined at $$x=1$$.

Next, for $$f(g(x)) = \frac{x}{x+1}$$

Since, the denominator can't be zero, x=-1 must also be excluded.

For $$g(f(x)) = \frac{x}{1-x}$$

$$\Rightarrow x=1$$ is not possible, which is already excluded.

So the values at which $$h(x)=f(g(x))+g(f(x))$$ is undefined are $${-1,\ \tfrac12,\ 1}$$