the answer is wrong and i cant find where's the mistake

In your second step,the denominator did not change; you wrote 3 instead of 2 as the numerator in the first frist fraction.And there is something else wrong in your work.

You can do it in this way,

  $${\frac{{\mathtt{2}}}{{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}^{{\mathtt{2}}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{3}}}{\left({{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{9}}\right)}} = {\mathtt{0}}$$

plus 3/(x^2-9) both side

$${\frac{{\mathtt{2}}}{{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}^{{\mathtt{2}}}}} = {\frac{{\mathtt{3}}}{\left({{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{9}}\right)}}$$

bot side times (x+3),

$${\frac{{\mathtt{2}}}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}} = {\frac{{\mathtt{3}}}{\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}}$$

$${\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{6}} = {\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{9}}$$

$${\mathtt{x}} = -{\mathtt{15}}$$

check: 2/(-15+3)^2-3/(x^2-9)=1/72-1/72=0 and x=-15 not equal to 3 or -3.

another way:

$${\frac{{\mathtt{2}}}{{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}^{{\mathtt{2}}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{3}}}{\left({{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{9}}\right)}} = {\mathtt{0}}$$

$${\frac{{\mathtt{2}}}{\left[\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right){\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)\right]}}{\mathtt{\,-\,}}{\frac{{\mathtt{3}}}{\left(\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right){\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right)\right)}} = {\mathtt{0}}$$

$${\frac{{\mathtt{2}}{\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}{\left[\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right){\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right){\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right)\right]}}{\mathtt{\,-\,}}{\frac{{\mathtt{3}}{\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}{\left[\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right){\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right){\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right)\right]}} = {\mathtt{0}}$$

both sides of the equation multiply by (x+3)^2(x-3) (x not equal 3 or -3)

$${\mathtt{2}}{\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right){\mathtt{\,-\,}}{\mathtt{3}}{\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right) = {\mathtt{0}}$$ 

$${\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{9}} = {\mathtt{0}}$$

$${\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{15}} = {\mathtt{0}}$$

$${\mathtt{x}} = -{\mathtt{15}}$$

thank you!dont understand how i could do such a stupid mistake

Hey things happen.

nobodys perfect but a few people (but that is a all out opinion, so really nobodys perfect)