I need to show, that $$\int v^2 \left( v_i v_j - v^2 \delta_{ij} \right) f_m d^3v = 0 ,$$ where $$f_m \propto \exp(-v^2), $$ is Maxwellian distributin. Actually, those indicies frustrates me, I know what is Kronecker delta. Should I take just two integrals: $$\int v^2 \left( v_x^2 - v^2 \right) f_m d^3v = 0 ,$$ $$\int v^2 v_x v_z f_m d^3v = 0 ,$$ and try to prove them?
So, I post it with error, sorry. First integral should be $$\int v^2 \left( v_i v_j - \frac13 v^2 \delta_{ij} \right) f_m d^3v = 0 ,$$
Recall that in this notation repeated indexes imply a sum, so your second expression – assuming you are integrating over $\mathbb R^3$– can be evaluated as follows: $$\int v^2 v_i v_j f_m d^3v = \int_{v_i>0} v^2 v_i v_j f_m d^3v + \int_{v_i<0} v^2 v_i v_j f_m d^3v$$ $$=\int_{v_i>0} v^2 v_i v_j f_m d^3v - \int_{v_i>0} v^2 v_i v_j f_m d^3v =0.$$
Edit (after OP's correction):
On the other hand, the first integrals add up to $$ \int v^2 \left( \sum_i v_i^2 - v^2/3 \right) f_m d^3v = \int v^2 \left( v^2 - v^2 \right) f_m d^3v = 0.$$