Answer:
[tex]\large\boxed{\dfrac{x+3}{x}}[/tex]
Step-by-step explanation:
[tex]\dfrac{\frac{x}{x-3}}{\frac{x^2}{x^2-9}}=\dfrac{x}{x-3}\div\dfrac{x^2}{x^2-9}=\dfrac{x}{x-3}\cdot\dfrac{x^2-9}{x^2}\\\\\text{cancel one x}\\\\=\dfrac{1}{x-3}\cdot\dfrac{x^2-3^2}{x}\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\dfrac{1}{x-3}\cdot\dfrac{(x-3)(x+3)}{x}\\\\\text{cancel (x-3)}\\\\=\dfrac{1}{1}\cdot\dfrac{x+3}{x}\\\\=\dfrac{x+3}{x}[/tex]
