Answer
$f^{-1}(x)=\dfrac{\sqrt{x}}{x}$
Work Step by Step
$f(x)=\dfrac{1}{x^{2}}$, $x\gt0$
Rewrite this expression as $y=\dfrac{1}{x^{2}}$ and solve for $x$:
$y=\dfrac{1}{x^{2}}$
Take $x^{2}$ to multiply the left side:
$x^{2}y=1$
Take $y$ to divide the right side:
$x^{2}=\dfrac{1}{y}$
Take the square root of both sides:
$\sqrt{x^{2}}=\sqrt{\dfrac{1}{y}}$
$x=\dfrac{1}{\sqrt{y}}$
Rationalize the denominator on the right side:
$x=\dfrac{1}{\sqrt{y}}\cdot\dfrac{\sqrt{y}}{\sqrt{y}}$
$x=\dfrac{\sqrt{y}}{y}$
Interchange $x$ and $y$:
$y=\dfrac{\sqrt{x}}{x}$
The inverse of the original function is $f^{-1}(x)=\dfrac{\sqrt{x}}{x}$
