1. **State the problem:** Differentiate the function $$P(x) = 18 - \frac{x^2}{x+3} - \sqrt{x+3}.$$\n\n2. **Rewrite the function for clarity:** $$P(x) = 18 - \frac{x^2}{x+3} - (x+3)^{\frac{1}{2}}.$$\n\n3. **Recall differentiation rules:**\n- Derivative of a constant is 0.\n- Use the quotient rule for $$\frac{x^2}{x+3}$$: If $$f(x) = \frac{u}{v},$$ then $$f'(x) = \frac{u'v - uv'}{v^2}.$$\n- Use the power rule for $$ (x+3)^{\frac{1}{2}} $$: $$\frac{d}{dx} (x+3)^n = n(x+3)^{n-1} \cdot 1$$ (chain rule with derivative of inside function 1).\n\n4. **Differentiate each term:**\n- Derivative of 18 is 0.\n- For $$\frac{x^2}{x+3}$$, let $$u = x^2$$ and $$v = x+3$$, so $$u' = 2x$$ and $$v' = 1$$.\nApply quotient rule:\n$$\frac{d}{dx} \frac{x^2}{x+3} = \frac{2x(x+3) - x^2(1)}{(x+3)^2}.$$\nSimplify numerator:\n$$2x(x+3) - x^2 = 2x^2 + 6x - x^2 = x^2 + 6x.$$\nSo derivative is:\n$$\frac{x^2 + 6x}{(x+3)^2}.$$\n\n- For $$-(x+3)^{\frac{1}{2}}$$, derivative is:\n$$- \frac{1}{2} (x+3)^{-\frac{1}{2}} \cdot 1 = - \frac{1}{2 \sqrt{x+3}}.$$\n\n5. **Combine all derivatives:**\n$$P'(x) = 0 - \frac{x^2 + 6x}{(x+3)^2} - \frac{1}{2 \sqrt{x+3}} = - \frac{x^2 + 6x}{(x+3)^2} - \frac{1}{2 \sqrt{x+3}}.$$\n\n6. **Check options:**\n- Option a: $$- \frac{x^3}{\sqrt{x+1}} - \sqrt{-3x+2}$$ (does not match)\n- Option b: $$2x + 12 \sqrt{x+3}$$ (does not match)\n- Option c: $$- \frac{x^2}{\sqrt{x+3}} + 2 \sqrt{x+3}$$ (does not match)\n- Option d: $$- \sqrt{2x+3}$$ (does not match)\n- Option e: None of the given options is correct.\n\n**Final answer:** Option e is correct because none of the given options matches the derivative.\n\n$$\boxed{P'(x) = - \frac{x^2 + 6x}{(x+3)^2} - \frac{1}{2 \sqrt{x+3}}}.$$