If $2f(x)+3f(\frac {1}{x})=\frac {4x^2+6}{x}$ and $f^{-1}(x)=1$ then find the value of $x$ If $2f(x)+3f(\frac {1}{x})=\frac {4x^2+6}{x}$ and $f^{-1}(x)=1$ then find the value of $x$.
My Attempt: 
$$2f(x)+3f(\frac {1}{x})=\frac {4x^2+6}{x}$$
$$2f(x)+3f(\frac {1}{x})=4x + \frac {6}{x}$$
At this point, I couldn't get other idea except comparing the corresponding terms. please provide any other method... 

You can solve the following system.
$2f(x)+3f\left(\frac{1}{x}\right)=\frac{4x^2+6}{x}$ and $2f\left(\frac{1}{x}\right)+3f(x)=\frac{\frac{4}{x^2}+6}{\frac{1}{x}}$,
which gives $f(x)=2x$.
