jmc

We begin by cross multiplying and then squaring both sides $$\begin{array}{rl}\frac{\sqrt{3x+5}}{\sqrt{6x+5}}& =\frac{\sqrt{5}}{3}\\ 3\sqrt{3x+5}& =\sqrt{5}\cdot \sqrt{6x+5}\\ {\left(3\sqrt{3x+5}\right)}^2& ={\left(\sqrt{5}\cdot \sqrt{6x+5}\right)}^2\\ 9(3x+5)& =5(6x+5)\\ 20& =3x\\ x& =\boxed{\frac{20}{3}}.\end{array}$$Checking, we see that this value of $x$ satisfies the original equation, so it is not an extraneous solution.