Suppose a differentiable function $$f(x)$$ satisfies the identity $$f(x + y) = f(x) + f(y) + xy^2 + x^2y$$, for all real $$x$$ and $$y$$. If $$\lim_{x \to 0}\frac{f(x)}{x} = 1$$, then $$f'(3)$$ is equal to __________

We are told that the differentiable function $$f(x)$$ satisfies the functional equation

$$f(x+y)=f(x)+f(y)+xy^{2}+x^{2}y$$ $$\text{for every real }x,y.$$

The presence of the mixed terms $$xy^{2}+x^{2}y$$ reminds us of the identity for the cube of a sum. Indeed, the algebraic formula

$$(x+y)^{3}=x^{3}+y^{3}+3x^{2}y+3xy^{2}$$

implies, after dividing by 3, that

$$\frac{(x+y)^{3}-x^{3}-y^{3}}{3}=x^{2}y+xy^{2}.$$

This observation suggests subtracting one-third of $$x^{3}$$ from $$f(x)$$ so that the mixed terms cancel. We therefore introduce a new function

$$g(x)=f(x)-\frac{x^{3}}{3}.$$

Now we compute $$g(x+y)$$:

$$\begin{aligned} g(x+y)&=f(x+y)-\frac{(x+y)^{3}}{3}\\[4pt] &=\Bigl[f(x)+f(y)+xy^{2}+x^{2}y\Bigr]-\frac{x^{3}+y^{3}+3x^{2}y+3xy^{2}}{3}\\[4pt] &=f(x)-\frac{x^{3}}{3}+f(y)-\frac{y^{3}}{3} +\underbrace{\bigl[xy^{2}+x^{2}y-\bigl(x^{2}y+xy^{2}\bigr)\bigr]}_{=\,0}\\[4pt] &=g(x)+g(y). \end{aligned}$$

Thus $$g(x)$$ is additive: $$g(x+y)=g(x)+g(y).$$ Because $$f$$ (hence $$g$$) is differentiable, the classical result for differentiable additive functions tells us that $$g(x)$$ must be linear. Hence there exists a constant $$k$$ such that

$$g(x)=kx\quad\text{for all }x.$$

Returning to $$f(x)$$, we therefore have

$$f(x)=g(x)+\frac{x^{3}}{3}=kx+\frac{x^{3}}{3}.$$

The problem also gives us the limit condition

$$\lim_{x\to 0}\frac{f(x)}{x}=1.$$

We substitute our expression for $$f(x)$$ into this limit:

$$\begin{aligned} \frac{f(x)}{x}&=\frac{kx+\dfrac{x^{3}}{3}}{x} =k+\frac{x^{2}}{3}. \end{aligned}$$

Taking the limit as $$x\to 0$$ simply removes the $$x^{2}$$ term, so

$$\lim_{x\to 0}\frac{f(x)}{x}=k=1.$$

Therefore $$k=1$$, and the explicit formula for $$f(x)$$ becomes

$$f(x)=x+\frac{x^{3}}{3}.$$

Because $$f$$ is differentiable, we differentiate term by term, using the basic rule $$\dfrac{d}{dx}(x^{n})=nx^{n-1}$$:

$$f'(x)=\frac{d}{dx}\Bigl(x\Bigr)+\frac{d}{dx}\Bigl(\frac{x^{3}}{3}\Bigr) =1+\frac{3x^{2}}{3}=1+x^{2}.$$

Finally, we evaluate the derivative at $$x=3$$:

$$f'(3)=1+3^{2}=1+9=10.$$

So, the answer is $$10$$.