How can I solve $f(t^2+u)=tf(t)+f(u)$ 
Solve $f(t^2+u)=tf(t)+f(u)$ on $\mathbb{R}$.

My solution is, if we take $u=t$, then $f(t^2+t)=(t+1)f(t)$ and let $g(x)=\frac{f(x)}{x}$, then $g(t^2+t)=g(t),(t\neq-1,0)$. It means 
$g(x)=0\quad or\quad g(x)=k,\quad k\in\mathbb{R}\tag{*}$
Therefore $\boxed{f(x)=0\quad or \quad f(x)=kx\quad\forall k,x\in\mathbb{R}}$. I'm not sure about $(*)$ is true.
 A: I want to succeed the answer given before me by using Cauchy's equation in an actual solution.
Let the given assertion be $P(t,u)$
$$P(1,0)\implies f(1)=f(1)+f(0) \Leftrightarrow f(0)=0$$
$$P(t,0) \implies f(t^2)=tf(t)+f(0)=tf(t)\tag{1}$$
Substituting $x=t^2$ and $(1)$ in $P(t,u)$
$$P(t,u) \implies f(t^2+u)=tf(t)+f(u)=f(t^2)+f(u)$$
$$\implies f(x+u)=f(x)+f(u) \tag{2}$$
But this holds for all $x \geq 0$, so
$$P(-t,0) \implies f(t^2)=-tf(-t)$$ 
Now combining that with $(1)$ we get
$$f(-t)=-f(t)$$
Or that $f$ is odd. So, substituting $x$ with $-x$ in $(2)$
$$f(u-x)=f(u)+f(-x)=f(u)-f(x) \text{  } \forall x \geq 0$$
or in other words
$$f(u+x)=f(u)+f(x) \text{  } \forall x \le 0$$
The fact that $x$ works $(2)$ implies that $-x$ also works in $(2)$ or the fact that $f$ is odd extends $(2)$ to work for all $x,u \in \mathbb{R}$.
Now, we're going to use $(2)$ many times by making partitioning sums inside functions (I mean $f(a_1+a_2+\dots+a_n)=f(a_1)+f(a_2)+\dots+f(a_n)$ for any positive integer $n$ and variables $a_i$)
$$P(t+1,0)\implies f(t^2+2t+1)=(t+1)f(t+1)$$
$$\Leftrightarrow f(t^2)+f(2t)+f(1)=(t+1)(f(t)+f(1))$$
and by $(1)$ and distributing the right-hand side,
$$\Leftrightarrow tf(t)+f(2t)+f(1)=tf(t)+tf(1)+f(t)+f(1)$$
$$\Leftrightarrow f(2t)=tf(1)+f(t) $$
and since $f(2t)=f(t+t)=f(t)+f(t)=2f(t)$
$$f(2t)=tf(1)+f(t) \Leftrightarrow 2f(t)=tf(1)+f(t)$$
$$\Leftrightarrow f(t)=tf(1)$$
Putting $f(1)=k$ and substituting it in the original equation, we get that it applies for any $k,t,u \in \mathbb{R}$, hence the conclusion. $\Box$
A: Given: $$f(t^2+u)=tf(t)+f(u) \tag{1}$$
Substituting $t = 1$ and $u = 0$ into (1), we get $$f(1)=f(1) + f(0) \Rightarrow f(0)=0\tag{2}$$
Substituting $u = 0$ and (2) into (1), we get $$f(t^2)=t f(t)\tag{3}$$
Substituting (3) and $v = t^2$ into (1), we get $$f(u+v)=f(u)+f(v) \tag{4}$$
Note that (4) holds only when $v = t^2 \ge 0$. To incorporate negative values, note that by substituting $t = -x$ in (3),
\begin{align}
 f((-x)^2) &= (-x) f(-x) \\
\Rightarrow f(x^2) &= -x f(-x) \\
\Rightarrow x f(x) &= -x f(-x) \qquad \textrm{from (3)} \\
\Rightarrow f(-x) &= -f(x)
\end{align}
Thus (4) holds for all $u,v\in\mathbb{R}$.
Equation (4) is Cauchy's functional equation and under some light assumptions (for example, continuity at one point), the only solution is $f(x) = cx$ for $c\in\mathbb{R}$.
See this SE post for a lot more detail on the solutions to Cauchy's functional equation.
A: Here is a proof without any use of Cauchy's functional equation.   First, as other users have observed: $f(0)=0$ and
$$f(t^2)=t\,f(t)$$
for all $t\in\mathbb{R}$.  Thus, if $x\geq 0$ and $y\in\mathbb{R}$, then
$$f(x+y)=f\big((\sqrt{x})^2+y\big)=\sqrt{x}\,f(\sqrt{x})+f(y)=f\big((\sqrt{x})^2\big)+f(y)=f(x)+f(y)\,.$$
Therefore, for any $t\in\mathbb{R}$, we have
$$f\big((t+1)^2\big)=f(t^2+2t+1)=f\big((t^2+t+1)+t\big)=f(t^2+t+1)+f(t)\,,$$
as $t^2+t+1>0$ for any $t\in\mathbb{R}$.  Next,
$$f(t^2+t+1)=f\big((t^2+1)+t\big)=f(t^2+1)+f(t)$$
because $t^2+1>0$ for all $t\in\mathbb{R}$.  Finally,
$$f(t^2+1)=f(t^2)+f(1)$$
because $t^2\geq 0$ for each $t\in\mathbb{R}$.  Consequently,
$$\begin{align}f\big((t+1)^2\big)&=f(t^2+t+1)+f(t)=\big(f(t^2+1)+f(t)\big)+f(t)\\&=\Big(\big(f(t^2)+f(1)\big)+f(t)\Big)+f(t)\,.\end{align}$$
Ergo,
$$f\big((t+1)^2\big)=f(t^2)+2\,f(t)+f(1)\,.\tag{*}$$
On the other hand,
$$f\big((t+1)^2\big)=(t+1)\,f(t+1)\,,$$
but
$$f(t+1)=f(1+t)=f(1)+f(t)\,,$$
as $1>0$.  Therefore,
$$f\big((t+1)^2\big)=(t+1)\,\big(f(1)+f(t)\big)=t\,f(t)+t\,f(1)+f(t)+f(1)\,.$$
Because $f(t^2)=t\,f(t)$, we conclude that
$$f\big((t+1)^2\big)=f(t^2)+t\,f(1)+f(t)+f(1)\,.\tag{#}$$
From (*) and (#), we conclude that
$$f(t)=t\,f(1)$$
for all $t\in\mathbb{R}$.
