Using the trace theorem $H^1(B(\overrightarrow{0},1))\hookrightarrow L^2(\partial B(\overrightarrow{0},1))$ and a change of variables, show that there exists $C_0>0$ such that for any $r>0$ and any $f\in H^1(B(\overrightarrow{0},1))$ the following holds: $$\int_{\partial B(\overrightarrow{0},1)}|f|^2\leq\displaystyle \frac{C_0}{r} \int_{B(\overrightarrow{0},1)}|f|^2dx+C_0r\int_{B(\overrightarrow{0},1)}|\nabla f|^2dx$$
I would like to use the divergence theorem to change the left side of the inequality, but I really do not how to start.
Oh, sorry, I make a mistake in the above inequality, that should be $$\int_{\partial B(\overrightarrow{0},r)}|f|^2\leq\displaystyle \frac{C_0}{r} \int_{B(\overrightarrow{0},r)}|f|^2dx+C_0r\int_{B(\overrightarrow{0},r)}|\nabla f|^2dx$$
The original statement is false.
Consider the constant function $f\equiv1$. It is in $H^1$ and $\nabla f\equiv0$. Then your estimate becomes $$ |\partial B(0,1)| \leq C_0r^{-1}|B(0,1)| $$ for all $r>0$. If this is true, then it is also true in the limit $r\to\infty$. Taking this limit says that the surface measure of the ball is zero.
The corrected statement is true.
Consider a function $f\in H^1(B(0,r))$ and define $g\in H^1(B(0,1))$ by $g(x)=f(rx)$. Then change the variables to turn integrals over $B(0,r)$ and its boundary into those for the unit ball. By continuity of the trace map for the unit ball we have the required estimate for $g$ when $r=1$. The three integrals get different scaling factors when you change variables, and this produces the coefficients $r^{\pm1}$.