Can I assume the continuity of this function ?? $$f(x+y) = f(x)+f(y)+ \vert{x}\vert y+xy^2, \forall x,y \in R \ \text{and} \ f'(0) = 0$$
To solve this function, and to apply any relation of calculus, do I need to assume that the given function is differentiable and continuous.
What I mean is , if I assume it to continouos and differentiable, I can get a function after solving.
Can there be a random non - differentiable or non continouos type of function that can also satisfy it??
Is there a way to tell that, this function will be continuous and differentiable before I start solving, just by looking
Basically what I am asking is , what things should I first take care of/ initial observations to make before getting started. (I am in highschool and don't have much experience in calculus. I read to learn rather than depend on a teacher, and I got a bit confused here a bit) 
Eg If it had been $$f(x+y) = f(x)+f(y), \forall x,y \in R \ \text{and} \ f'(0) = 1$$
Can I take the derivative wrt x, keeping y constant and say $$f'(x+y) = f'(x)$$ 
And put x=0 and take y as a changing parameter and write $$f'(y) = f'(0)=1$$ and thus integrate it again to get $$f(x)=x+c$$ and put x and y = 0 , to get  $$f(0) = 2f(0)$$, so $f(0)=0$ , thus c=0 and I get $$f(x)=x$$
This what I was taught. But it just feels wrong.... Neither is the continuity of function given , nor its differentiablity given... Also first we sat y is constant, then take it as changing parameter(partial derivatives is not in our syllabus , so I suspect the teacher wanted to take partial derivative, but I'm not sure. I have also not read a lot about them, but know that they exist)
Also Can other solutions also satisfy this equation( Not this eg in particular , but in general for any functional equation, any extra function that might come satisfy it, but will not be deduced via this method....) 
 A: "Just by looking", you can only say that $f'(0)$ exists, hence that $f$ is continuous and once differentiable at $x=0$. I suppose it will follow from this little spot that $f$ is continuous throughout. But it cannot be differentiable: Apply $\frac{\mathrm d}{\mathrm d x}$ at $(0,1)$ and you would obtain a derivative of $|x|$.
Actually, if $f$ is a solution then also
$$f(y+x)=f(y)+f(x)+|y|x+yx^2 $$
for all $x,y\in\Bbb R$. So it would follow that
$$ |x|y+xy^2=|y|x+yx^2$$
for all $x,y\in \Bbb R$, which is absurd. With or without any assumptions of continuity, no solution on all of $\Bbb R$ or even on an interval around $0$ can exist.
A: The function is continuous at $0$, because it is differentiable at $0$. We'd like to prove that it is continuous everywhere, i.e. that $$\lim_{y\to0}f(x+y)-f(x)=0$$
Since $f(x+y)-f(x)=f(y)+\lvert x\rvert y+xy^2$ and $f$ is continuous at $0$, $$\lim_{x\to0} f(x+y)-f(y)=f(0)$$
And $f(0)=f(0+0)=2f(0)$. Therefore $f(0)=0$, proving continuity.
We'd like to prove differentiability everywhere, i.e. that $\lim_{y\to 0}\frac{f(x+y)-f(y)}{y}$ exists. Again, $$\frac{f(x+y)-f(y)}{y}=\frac{f(y)}y+\lvert x\rvert+xy=\frac{f(y)-f(0)}y+\lvert x\rvert+xy$$
Therefore $\lim_{y\to 0}\frac{f(x+y)-f(y)}{y}=f'(0)+\lvert x\rvert=\lvert x\rvert$.
We've just proved that $f(0)=0$ and $f'(x)=\lvert x\rvert$. Since $f'$ is continuous, $$f(x)=\int_0^x \lvert t\rvert\,dt=\frac12x\lvert x\rvert$$
Finally, does this $f$ satisfy the original functional equation? $$\frac12(x+y)\lvert x+y\rvert-\frac12x\lvert x\rvert-\frac12y\lvert y\rvert-\lvert x\rvert y-xy^2\stackrel?= 0$$
And the answer just so happens to be NO. Therefore there is no such function.
