If $3c$ is a perfect square then why must $c$ be of form $3x^2$ for some $x$? ( $x,c$ integers) For example $3c=36$,
then $36=3*4^2$ so $x=4$.

Primes (such as $3$) have the property that if they divide a product, they divide (at least) one of the factors: If $p\mid ab$ then $p\mid a$ or $p\mid b$. 
Here we take $p=3$ and from the assumption $3c=m^2=m\cdot m$ conclude that $3$ divides one of the two factors on the right, that is $3\mid m$ at any rate. So $m=3x$ for some integer $x$, hence $3x=m^2$ turns into $3c=9x^2$, i.e. $c=3x^2$.
