Show that $f: \mathbb N \to \mathbb N$, $f(x)=x^2$ is not onto To begin, the definition of an onto (surjective) function is as follows.
A function $\phi$ from $A$ to $B$ is surjective if for each for each $b$ in $B$, there exists at least one $a$ in $A$ such that $\phi(a)=b$. 
Let $f:\mathbb{N}\to \mathbb{N}$ and $x\in \mathbb{N}$ and $y\in \mathbb{N}$. 
We have that $y=f(x)=x^2$, which gives $x=\pm  \sqrt y$. 
If $x=- \sqrt y$, we have that $y=x^2$. And if $x = \sqrt y$, we have that $y=x^2$. 
This shows there is at least one element $x$ that is the pre-image of an element $y$. I don't see how this function can be proven to be not-onto.

Take $y$ to be a square-free natural number, then there isn't any natural number $x$ such that $f(x)=y$ (For example, can you find any natural number such that square of it is equal to 2?). So $f$ can't be onto.
