The function given is:

$f(x) = \frac{3 - xe^x}{x + e^x}$

To differentiate this, we will apply the quotient rule, which is used when differentiating a ratio of two functions. The quotient rule states:

$\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x) u'(x) - u(x) v'(x)}{(v(x))^2}$

Here, $u(x) = 3 - xe^x$ and $v(x) = x + e^x$.

Step-by-step differentiation:

1. Differentiate $u(x) = 3 - xe^x$: $u'(x) = -e^x - xe^x$ (This uses the product rule for differentiating $xe^x$, and the derivative of 3 is 0.)

2. Differentiate $v(x) = x + e^x$: $v'(x) = 1 + e^x$

3. Apply the quotient rule: $f'(x) = \frac{(x + e^x)(-e^x - xe^x) - (3 - xe^x)(1 + e^x)}{(x + e^x)^2}$

4. Simplify the numerator:

   Distribute both terms in the numerator:

   $(x + e^x)(-e^x - xe^x) = -xe^{2x} - x e^x - e^{2x} - e^x$ $(3 - xe^x)(1 + e^x) = 3 + 3e^x - xe^x - xe^{2x}$

   Now subtract the two expressions:

   $\left(-xe^{2x} - x e^x - e^{2x} - e^x\right) - \left(3 + 3e^x - xe^x - xe^{2x}\right)$

   Simplify further:

   $-xe^{2x} - e^{2x} - x e^x - e^x - 3 - 3e^x + xe^x + xe^{2x}$

   Combine like terms:

   $-e^{2x} - 4e^x - 3$

5. Final derivative expression: $f'(x) = \frac{-e^{2x} - 4e^x - 3}{(x + e^x)^2}$

This is the derivative of the given function.

Would you like more details or further explanations?

Here are 5 related questions you could explore:

1. How would you differentiate the same function using logarithmic differentiation?
2. Can you solve for critical points of $f(x)$ using the derivative?
3. What is the significance of the quotient rule in calculus?
4. How does the presence of $e^x$ influence the behavior of the function as $x$ approaches infinity?
5. How can we verify the result of this derivative using an alternative method?

Tip: When simplifying expressions, be careful to track negative signs and distribution to avoid common algebraic mistakes.