Determine all $f:\mathbb{R}\to\mathbb{R}$ such that $(x+y)\,\big(f(x)-f(y)\big)=f(x^2)-f(y^2)$. 
Determine all $f:\mathbb{R}\to\mathbb{R}$ such that $$(x+y)\,\big(f(x)-f(y)\big)=f(x^2)-f(y^2)\,.$$ 

Help pls! Thank you.

Edit (by Batominovski).
See the edit history of this question.  Originally, the functional equation $$(x+y)\,\big(f(x){\color{red}-}f(y)\big)=f(x^2)-f(y^2)$$ was in the title, but the body asked about $$(x+y)\,\big(f(x){\color{red}+}f(y)\big)=f(x^2)-f(y^2).$$  The OP has not clarified which functional equation is the real question.  
Note : The functional equation that Mr. Batominovski solved first was the real question. I’m sorry for mistake in typing. Thank you for helping! 
 A: Let $f:\mathbb{R}\to\mathbb{R}$ satisfy 
$$(x+y)\,\big(f(x)-f(y)\big)=f(x^2)-f(y^2)\,.$$
for all $x,y\in\mathbb{R}$.  Let $c:=f(0)$.  Plugging $y:=0$ into the original functional equation yields
$$f(x^2)=x\,\big(f(x)-c\big)$$
for each $x\in\mathbb{R}$.  Using this result in the original functional equation, we obtain
$$(x+y)\,\big(f(x)-f(y)\big)=x\,\big(f(x)-c\big)-y\,\big(f(y)-c\big)$$
for all $x,y\in\mathbb{R}$.  After some algebraic manipulations, we get
$$y\,\big(f(x)-c\big)=x\,\big(f(y)-c\big)$$
for all $x,y\in\mathbb{R}$.  Hence, for any real numbers $x,y\neq 0$,
$$\frac{f(x)-c}{x}=\frac{f(y)-c}{y}\,.$$
Thus, there exists a constant $k$ such that
$$\frac{f(x)-c}{x}=k$$
for every real number $x\neq 0$.  This shows that
$$f(x)=kx+c\text{
for all real numbers $x\neq 0$}\,.$$  However, $f(0)=c$ implies that $$f(x)=kx+c\text{ for all $x\in\mathbb{R}$}\,.$$  It is easy to see that any function of the form above (i.e., $k$ and $c$ can be arbitrary real numbers) is a solution.

Let $f:\mathbb{R}\to\mathbb{R}$ satisfy
$$(x+y)\,\big(f(x)+f(y)\big)=f(x^2)-f(y^2)$$
for all $x,y\in\mathbb{R}$.  Then, by letting $y:=x$ in the original functional equation, we obtain
$$4x\,f(x)=0$$
for all $x\in\mathbb{R}$.  In particular, this shows that $f(x)=0$ for every $x\in\mathbb{R}_{\neq 0}$.  Now, by letting $x:=1$ and $y:=0$ in the original functional equation and using $f(1)=0$, we have
$$f(0)=-f(0)\,.$$
Therefore, $f(0)=0$.  Ergo,
$$f(x)=0\text{ for all }x\in\mathbb{R}\,,$$
which is easily seen to be a solution.
Edit.  The solution for another functional equation was added, because Tavish remarked that the functional equation was changed.  The OP wrote two different functional equations.  When I edited the question to fix grammatical mistakes and to improve the format, I inadvertently changed one functional equation the OP had written to another.  The OP has not yet clarified which functional equation is the real question.
A: From
$$
\cases{
(x-y)(f(x)-f(-y))=f(x^2)-f(y^2)\\
-(x-y)(f(-x)-f(y))=f(x^2)-f(y^2)
}\Rightarrow f(x)+f(-x)=f(y)+f(-y)\Rightarrow f(-x)+f(x) = c
$$
now
$$
\cases{
x+y=\frac{f(x^2)-f(y^2)}{f(x)-f(y)}\\
x-y=\frac{f(x^2)-f(y^2)}{f(x)+f(y)-c}
}\Rightarrow 
\cases{
f(x) = \frac c2+\frac{f(x^2)-f(y^2)}{x^2-y^2}x\\
f(y) = \frac c2+\frac{f(x^2)-f(y^2)}{x^2-y^2}y
}\Rightarrow f(x) = k x+c_0
$$
