1. We are given

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

Also,

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

We need to find $f'(3)$.

2. First, simplify the extra term:

$xy^2+x^2y=xy(x+y).$

So the functional equation is

$f(x+y)=f(x)+f(y)+xy(x+y).$

This suggests subtracting a cubic term, because

$(x+y)^3=x^3+y^3+3xy(x+y).$

Hence,

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

Therefore, if we define

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

then

\begin{align*} g(x+y) &= f(x+y)-\frac{(x+y)^3}{3}\ &= f(x)+f(y)+xy(x+y)-\left(\frac{x^3}{3}+\frac{y^3}{3}+xy(x+y)\right)\ &= \left(f(x)-\frac{x^3}{3}\right)+\left(f(y)-\frac{y^3}{3}\right)\ &= g(x)+g(y). \end{align*}

So $g$ satisfies Cauchy’s additive equation:

$g(x+y)=g(x)+g(y).$

3. Since $f$ is differentiable, $g(x)=f(x)-x^3/3$ is also differentiable, hence continuous. A continuous additive function must be linear, so

$g(x)=cx$

for some constant $c$.

Thus,

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

4. Use the given limit:

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

Now,

$\frac{f(x)}{x}=\frac{cx+x^3/3}{x}=c+\frac{x^2}{3}.$

Taking limit as $x\to 0$,

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

So,

$c=1.$

Hence,

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

5. Differentiate:

$f'(x)=1+x^2.$

Therefore,

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

6. Final answer:

$\boxed{10}$

This matches the stored correct answer.