Play with functions

Given $f(1)=\frac { 1 }{ 2 } \cdot (f\left( y \right) +f\left( \frac { 1 }{ y } \right) )$ and $f\left( 1 \right) = 0$ we have $f\left( y \right) =-f\left( \frac { 1 }{ y } \right) .................eq. 1$ . Now $f'(x)=\frac { 1 }{ 2 } [f'(xy)\cdot y+\frac { f'(\frac { x }{ y } ) }{ y } ]$ . Given $f'(1)=2\Longrightarrow yf'(y)+\frac { f'(\frac { 1 }{ y } ) }{ y } = 4 .....................eq. 2$ . Differentiating $eq. 1$ we have $f'(y)=\frac { f'(\frac { 1 }{ y } ) }{ { y }^{ 2 } }$ . Also $eq.2 \Longrightarrow f'(\frac { 1 }{ y } )=2y$ and $f'(y)=\frac { 2 }{ y }$ . So $f'(3)=\frac { y\cdot \frac { 2 }{ 3y } +\frac { 1 }{ y } \cdot \frac { 2y }{ 3 } }{ 2 } = \frac { 2 }{ 3 }$

The moment I saw the question, I tried guessing the function and lo! logarithmic function perfectly satisfies the conditions.

The work ahead is now extremely easy.

Let $f(x)=k\times log_{e}x$ , where $k$ is a constant.

Now,

$f^{'}(x)$ $=\frac{k}{x}$ .

But, $f^{'}(1)$ $=2$ implying that $k=2$ .

Thus, $f^{'}(3)$ $=\frac{2}{3}$ and we are done.

P.S.

I know this is not a rigorous solution so I would be glad if someone could add one, @abdulmuttalib lokhandwala ? Also, the euro symbol in the question looks odd, for the symbol of 'belongs to', use \in