Hint: use telescopy, i.e. a very simple telescopic induction proof yields
$\rm\ f(n)\ =\ \sum_{i\: =\: 1}^n\:\ (3i-1)^2\ \ \iff\ \ \ f(n) - f(n-1)\ =\ (3n-1)^2,\quad\ f(0) = 0$
So the proof reduces to showing that $\rm\:f(n) = n(6n^2 + 3n - 1)/2\:$ satisfies the RHS equations. Clearly $\rm\:f(0) = 0$. The other equality is between two quadratics, so to prove it we need only show they are equal at $3$ points, say $\rm\:n = 0,1,2.\:$ Since $\rm\:f(-1)=-1,\:f(0)=0,\:f(1)=4,\:f(2)=29$
$\rm\qquad\qquad n=0:\quad f(0)-f(-1)= 0-(-1) =\: (3\cdot 0-1)^2$
$\rm\qquad\qquad n=1:\quad f(1)\ -\ f(0)\:= 4\ \ -\ \ 0\ =\:\ (3\cdot 1 -1)^2$
$\rm\qquad\qquad n=2:\quad f(2)\ -\ f(1)\:= 29\ -\ 4\: =\:\ (3\cdot 2 -1)^2\quad$ QED
Note that the above method yields a simple mechanical algorithm for constructing such proofs.