Find my x

2f( $\frac{x}{1-x}$ ) + f( $\frac{1}{x-1}$ ) = $\frac{3}{x-1}$ ... (a)

Replace $x$ by $\frac{1}{x}$ , we get:

2f( $\frac{1/x}{1-1/x}$ ) + f( $\frac{1}{1/x-1}$ ) = $\frac{3}{1/x-1}$

$\implies$ 2f( $\frac{1}{x-1}$ ) + f( $\frac{x}{1-x}$ ) = $\frac{3x}{1-x}$

Now multiply both sides by 2;

$\implies$ 4f( $\frac{1}{x-1}$ ) + 2f( $\frac{x}{1-x}$ ) = $\frac{6x}{1-x}$ ... (b)

Perform the operation (b) - (a);

4f( $\frac{1}{x-1}$ ) + 2f( $\frac{x}{1-x}$ ) - 2f( $\frac{x}{1-x}$ ) - f( $\frac{1}{x-1}$ ) = $\frac{6x}{1-x}$ - $\frac{3}{x-1}$

$\implies$ 3f( $\frac{1}{x-1}$ ) = $\frac{6x}{1-x}$ + $\frac{3}{1-x}$

$\implies$ 3f( $\frac{1}{x-1}$ ) = $\frac{6x+3}{1-x}$

$\implies$ f( $\frac{1}{x-1}$ ) = $\frac{2x+1}{1-x}$

Put $x$ = 0;

f( $\frac{1}{0-1}$ ) = $\frac{2(0)+1}{1-0}$

f(-1) = 1