jmc

From the factorization $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc),$$we know that $a^3+b^3+c^3=3abc.$

Since $a+b+c=0,$ $c=-a-b,$ so $$\begin{array}{rl}{a}^5+b^5+c^5& =a^5+b^5-(a+b{)}^5\\ & =-5a^{4}b-10a^3{b}^2-10a^2{b}^3-5a{b}^4\\ & =-5ab(a^3+2a^{2}b+2a{b}^2+b^3)\\ & =-5ab[(a^3+b^3)+(2a^{2}b+2a{b}^2)]\\ & =-5ab[(a+b)(a^2-ab+b^2)+2ab(a+b)]\\ & =-5ab(a+b)(a^2+ab+b^2)\\ & =5abc(a^2+ab+b^2),\end{array}$$so $$3abc=5abc(a^2+ab+b^2).$$Since $a,$ $b,$ $c$ are all nonzero, we can write $$a^2+ab+b^2=\frac{3}{5}.$$Hence, $$\begin{array}{rl}{a}^2+b^2+c^2& =a^2+b^2+(a+b{)}^2\\ & =a^2+b^2+a^2+2ab+b^2\\ & =2a^2+2ab+2b^2\\ & =2(a^2+ab+b^2)=\boxed{\frac{6}{5}}.\end{array}$$