You may prove convexity from first principle. $f(x)$ is said to be strictly convex if and only if $f(px+qy)<pf(x)+qf(y)$ for any $x\not=y$ and for every $0<p<1 \ (q=1-p)$. This can be proved by verifying that
where $\|v\|$ denotes the length (i.e. Euclidean norm) of a vector $v$.
To find the minimum of $f$, note that $f(x)=x^Tx=\|x\|^2\ge0$ and $f(x)=0$ only when $x=0$. Hence the absolute minimum of $f$ occurs at the $x=0$. Calculus is not of much use here, because it can only prove that a certain point is a local minimum, but you are asked to find the absolute minima of $f$. But anyway, since you have $f'(x)=2x^T$, setting $f'(x)=0$ would give you back the critical point $x=0$. To show that $x=0$ is indeed an absolute minimum, you still need to argue that $f(x)\ge f(0)=0$ for every $x$.
Edit: It's worth mentioning (thanks to Dominic Michaelis) that every local minimum of a convex function is a global minimum, but in general, calculus is only helpful for screening out local minima among critical points. Extra work is often required to locate global minima.