Answer
The solution set is $\left\{-6, -1\right\}$.
Work Step by Step
Factor each denominator completely to obtain:
$\dfrac{2x}{x-3}+\dfrac{6-2x}{(x-3)(x+3)}=\dfrac{x}{x+3}$
The LCD is $(x-3)(x+3)$. Multiply this LCD to both sides of the equation to get rid of the denominators:
\begin{array}{ccc}
\require{cancel}
&(x-3)(x+3) \cdot \left[\dfrac{2x}{x-3}+\dfrac{6-2x}{(x-3)(x+3)}\right]&=&\dfrac{x}{x+3}\cdot (x-3)(x+3)
&\\&(x-3)(x+3) \cdot \dfrac{2x}{x-3}+(x-3)(x+3)\cdot \dfrac{6-2x}{(x-3)(x+3)}&=&\dfrac{x}{\cancel{x+3}}\cdot (x-3)\cancel{(x+3)}
&\\&\cancel{(x-3)}(x+3) \cdot \dfrac{2x}{\cancel{x-3}}+\cancel{(x-3)(x+3)}\cdot \dfrac{6-2x}{\cancel{(x-3)(x+3)}}&=&x(x-3)
\\&(x+3)(2x)+6-2x&=&x^2-3x
\\&2x^2+6x+6-2x&=&x^2-3x
\\&2x^2+4x+6&=&x^2-3x
\\&2x^2+4x+6-x^2+3x&=&0
\\&x^2+7x+6&=&0
\end{array}
Factor the trinomial:
$(x+6)(x+1)=0$
Use the Zero Factor Property by equating each factor to zero, then solve each equation to obtain:
\begin{array}{ccc}
&x+6=0 &\text{or} &x+1=0
\\&x=-6 &\text{or} &x=-1
\end{array}
