The problem is: I have a continuous $f: \mathbb R_+ \to \mathbb R$ which satisfies $f(xy)=xf(y)+yf(x)$ for all $x,y \in \mathbb R_+$. I want to prove that $f$ is differentiable.
The problem is equivalent to proving that $f'(1)$ exist. I have no idea how to go on.
On
Using integrals the job is easy. Since $f$ is continuous the anti-derivarive of $f$, say $F$ exists. Now integrate the functional equation with respect to $y$ on interval $[1,2]$ to get $$\frac{F(2x)-F(x)}{x}=x(F(2)-F(1))+ \frac{3}{2}f(x)$$ Since $F'$ exists it now follows that $f'(x) $ exists for all $x\in(0,\infty)$.
The problem is difficult if we stick to differential calculus only. I have managed to make some progress to solve the problem using methods of differential calculus and I present the same below.
Dividing by $xy$ we get the functional equation $$\frac{f(xy)}{xy} = \frac{f(x)}{x} + \frac{f(y)}{y}\tag{1}$$ and if we put $g(x) = f(x)/x$ then $$g(xy) = g(x) + g(y)\tag{2}$$ We prove the following theorem:
Theorem: If $g:\mathbb{R}^{+}\to\mathbb{R}$ is continuous at a point $c \in \mathbb{R}^{+}$ and satisfies $g(xy) = g(x) + g(y)$ for all $x, y \in \mathbb{R}^{+}$ then $g(x)$ is differentiable for all $x \in \mathbb{R}^{+}$ and $g'(x) = g'(1)/x$.
First we prove that $g$ is continuous on $\mathbb{R}^{+}$. Let $x > 0$ and $x \neq c$. Then we have $$\lim_{h \to 0}g(x + h) = \lim_{h \to 0}g(x/c)+g (c + ch/x) = g(x/c)+g(c) = g(x)$$ From the functional equation the following properties are obvious $$g(1) = 0, g(1/x) = -g(x), g(x/y) = g(x) - g(y), g(x^{q}) = qg(x), q\in \mathbb{Q}\tag{3}$$ Next we prove that $x = 1$ is the only solution for $g(x) = 0$. More precisely we prove that
If a continuous function $g:\mathbb{R}^{+}\to\mathbb{R}$ satisfies functional equation $(2)$ and there is a number $k \neq 1$ such that $g(k) = 0$ then $g(x) = 0$ for all $x\in\mathbb{R}^{+}$.
Let us suppose that $k \neq 1$ and $g(k) = 0$ and then by $(3)$ we have $g(k^{q}) = 0$ for all $q \in \mathbb{Q}$. Now it can be proved with some effort that every positive real number is an accumulation point of set $A = \{k^{q}: q\in \mathbb{Q}\}$ or in other words if we take any positive real number $x$ then every neighborhood of $x$ contains points of the form $k^{q}$. It now follows by sign preserving nature of continuous functions that $g(x) = 0$ for all positive $x$.
So one solution of the functional equation is the constant $0$ function and apart from this trivial solution any other $g$ is such that $g(x) = 0\Rightarrow x = 1$. From the relation $g(x/y) = g(x) - g(y)$ it then follows that $g$ is one-one function. We further note that any continuous one one function is strictly monotone (this is an easy consequence of intermediate value property of continuous functions) and hence $g$ is strictly monotone.
It follows that if $k \neq 1$ then $g(k)\neq 0$ and $g(k^{n}) = ng(k)$ for all integers $n$ so that function $g$ can take arbitrarily large negative and positive values. By intermediate value property it follows that range of $g$ is $\mathbb{R}$ and hence $g$ is one-one onto function.
Next we prove that if $g'(1)$ exists then $g'(x) = g'(1)/x$. Clearly we have \begin{align} g'(x) &= \lim_{h \to 0}\frac{g(x + h) - g(x)}{h}\notag\\ &= \lim_{h \to 0}\frac{g(1 + h/x)}{h}\notag\\ &= \lim_{h \to 0}\frac{g(1 + h/x) - g(1)}{h/x}\cdot\frac{1}{x}\notag\\ &= \frac{g'(1)}{x}\notag \end{align} We next need to show that $g'(1)$ exists or more explicitly the following limit exists $$\lim_{x \to 1}\frac{g(x)}{x - 1} = g'(1)\tag{4}$$ I have not been able to prove the existence of this limit in a direct manner. Rather I use the fact that $g$ is invertible and thus there is a function $h:\mathbb{R}\to\mathbb{R}^{+}$ such that $g(h(x)) = x, x\in \mathbb{R}$ and $h(g(x)) = x, x \in \mathbb{R}^{+}$. Moreover because of equation $(2)$ the function $h$ satisfies the functional equation $$h(x+y) = h(x)h(y)\tag{5}$$ and it has already been established that under these conditions the limit $$\lim_{x \to 0}\frac{h(x) - 1}{x}\tag{6}$$ exists and is non-zero unless $h$ is constant. Putting $h(x) - 1 = t$ we see that $x = g(1 + t)$ and as $x \to 0, t \to 0$ and thus $$\lim_{t \to 0}\frac{g(1 + t)}{t}$$ exists and thus the limit $g'(1)$ in equation $(4)$ exists.
It is now clear that $f(x) = xg(x)$ is differentiable for all $x > 0$.
In the above we have only discussed that if a function $g$ satisfies the equation $(2)$ and is continuous then it is differentiable also. With slightly more effort we can show that such functions actually exist. One way is to integrate the equation $g'(x) = g'(1)/x$ to get $$g(x) = g'(1)\int_{1}^{x}\frac{dt}{t}\tag{7}$$ so that any multiple of the above integral is a solution for $(2)$. There is another way to find an expression for $g$. Let $x\neq 1$ and then we know that $x^{1/n}\neq 1$ for all positive integers $n$ and also $x^{1/n}\to 1$ as $n\to\infty $. We then have $$g(x) =ng(x^{1/n})=n(x^{1/n}-1)\cdot\frac{g(x^{1/n})}{x^{1/n}-1}$$ Taking limits as $n\to\infty$ and noting that fraction on right in above equation tends to $g'(1)$ we can see that $$g(x) = g'(1)\lim_{n\to\infty}n(x^{1/n}-1)\tag{8}$$ Thus the solution $g$ is always a constant multiple of the limit in above equation. Both the integral in $(7)$ and the limit in $(8)$ are equivalent expressions for the logarithm function conventionally denoted by $\log x$ and the theory of this function can be developed independently using either the integral or the limit approach. The current answer shows that using functional equation $(2)$ and assumption of continuity we can discover a very useful and important function called logarithm.
Divide the functional equation by $xy$ to get
$$\frac{f(xy)}{xy}=\frac{f(y)}y+\frac{f(x)}x$$
Set $g(x)=\frac{f(x)}x$ so that
$$g(xy)=g(y)+g(x)$$
This is actually well known, and if $g(x)$ is continuous, then the solution is given by
$$g(x)=a\log(x)$$
which implies
$$f(x)=ax\log(x)$$