If 2f(x)-3f(1/x)=x^2, x is not equal to zero, then f(2) is equal to

$2f(x)-3f(\frac{1}{x})=x^2$   ...(1)

so as $x→\frac{1}{x}$

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

eq. (1) × 2 + eq. (2) × 3

$4f(x)-6f(\frac{1}{x})=2x^2 + 6f(\frac{1}{x})-9f(x)=\frac{3}{x^2}$

$-5f(x)=2x^2+\frac{3}{x^2}$

at $x=2$

$⇒-5f(2)=8+\frac{3}{4}⇒-\frac{7}{4}$