= \(\dfrac{x}{x-3}-\dfrac{x^2+3x}{2x+3}.\left(\dfrac{\left(x+3\right).\left(x+3\right)}{x^3-9x}-\dfrac{x^2}{x^3-9x}\right)\\ \)
= \(\dfrac{x}{x-3}-\dfrac{x\left(x+3\right)}{2x+3}.\dfrac{\left(2x+3\right).3}{x\left(x+3\right).\left(x-3\right)}\)
= \(\dfrac{x}{x-3}-\dfrac{3}{x-3}\)
= \(\dfrac{x-3}{x-3}\)
= 1