Prove that $f^{-1}:\mathbb{N} \rightarrow \mathbb{N}$ is partial recursive
You can partially compute $f^{-1}(y)$ as follows:


  for $x$ from $0$ to $\infty$:
  
  &emsp; if $f(x) = y$ then return $x$


If you don't like for loops, you can use the following implementation:


  
  $x \gets 0$
  if $f(x) = y$ then return $x$
  $x \gets x + 1$
  goto step 2