Find the functions satisfying $ \frac { f ( a b ) } { a b } = \frac { f ( a + b ) } { a + b } \cdot \frac { f ( a - b ) } { a - b } $ 
Find all the functions $ f : \mathbb R \to\mathbb R $ such that
$$ \frac { f ( a b ) } { a b } = \frac { f ( a + b ) } { a + b } \cdot \frac { f ( a - b ) } { a - b } \text . $$

I think $ f ( x ) = x $, but I don't know how to get it.

First of all, note that you have $ f : \mathbb R \to \mathbb R $, and in particular $ 0 $ is in the domain of $ f $. But the functional equation
$$ \frac { f ( x y ) } { x y } = \frac { f ( x + y ) } { x + y } \cdot \frac { f ( x - y ) } { x - y } \tag 0 \label 0 $$
can only hold when $ x , y \ne 0 $ and $ x \ne \pm y $. That means \eqref{0} doesn't tell you anything about $ f ( 0 ) $, and given any solution $ f $ of \eqref{0}, you can change the value of $ f ( 0 ) $, and still get a solution; i.e. any function $ \tilde f : \mathbb R \to \mathbb R $ with $ \tilde f ( x ) = f ( x ) $ for all $ x \ne 0 $ is a solution. Thus the better question is to ask for all $ f : \mathbb R \setminus \{ 0 \} \to \mathbb R $ satisfying \eqref{0} for all $ x , y \in \mathbb R \setminus \{ 0 \} $ with $ x \ne \pm y $. Then, defining $ g : \mathbb R \setminus \{ 0 \} \to \mathbb R $ with $ g ( x ) = \frac { f ( x ) } x $, \eqref{0} gives
$$ g ( x y ) = g ( x + y ) g ( x - y ) \tag 1 \label 1 $$
for all nonzero $ x $ and $ y $ with $ x \ne \pm y $. Conversely, any $ g : \mathbb R \setminus \{ 0 \} \to \mathbb R $ satisfying \eqref{1} gives rise to an $ f : \mathbb R \setminus \{ 0 \} \to \mathbb R $ defined with $ f ( x ) = x g ( x ) $, which satisfies \eqref{0}. So we will try to solve this equivalent problem. It's straightforward to check that the constant functions $ g ( x ) = 0 $ and $ g ( x ) = 1 $ satisfy \eqref{1} (equivalently, the constant zero function and the identity function both satisfy \eqref{0}). We will show that these are the only solutions.
Setting $ y = 1 $ and then $ y = - 1 $ in \eqref{1} we get
$$ g ( x ) = g ( x + 1 ) g ( x - 1 ) = g ( x - 1 ) g ( x + 1 ) = g ( - x ) $$
for all nonzero $ x $ with $ x \ne \pm 1 $. Similarly, substituting $ \frac x 2 $ for $ x $ in \eqref{1} and once setting $ y = 2 $ and once $ y = - 2 $, we get
$$ g ( x ) = g \left( \frac x 2 + 2 \right) g \left( \frac x 2 - 2 \right) = g \left( \frac x 2 - 2 \right) g \left( \frac x 2 + 2 \right) = g ( - x ) $$
for all nonzero $ x $ with $ x \ne \pm 4 $. Thus we have
$$ g ( - x ) = g ( x ) \tag 2 \label 2 $$
for all nonzero $ x $.
Now, suppose there is a nonzero $ a $ with $ g ( a ) = 0 $. Without loss of generality, we can assume that $ a > 0 $, because if $ a < 0 $, by \eqref{2} we have $ g ( - a ) = 0 $ while at the same time $ - a > 0 $. Letting $ x = y + a $ in \eqref{1} we get $ g \big( y ( y + a ) \big) = 0 $ for all positive $ y $ (note that since $ a > 0 $, thus $ 0 \ne y + a \ne \pm y $). This means that we have $ g ( x ) = 0 $ for all positive $ x $, as when $ x > 0 $ there is a positive $ y $, namely $ y = \frac { - a + \sqrt { a ^ 2 + 4 x } } 2 $, such that $ y ( y + a ) = x $. Using \eqref{2}, this can be generalized for negative $ x $ as well, which means that $ g $ is the constant zero function, one of the mentioned solutions.
Finally, consider the other case where for all nonzero $ x $, $ g ( x ) $ is not equal to $ 0 $. Suppose that $ y $ is negative. Then we have $ 0 \ne \frac y { y - 1 } \ne \pm y $, and we can let $ x = \frac y { y - 1 } $ in \eqref{1}. In that case, we have $ x y = x + y = \frac { y ^ 2 } { y - 1 } $, and by the fact $ g \left( \frac { y ^ 2 } { y - 1 } \right) $ is not equal to $ 0 $, we find out that $ g ( x - y ) = g \left( \frac { y ^ 2 - 2 y } { y - 1 } \right) = 1 $. This means that we have $ g ( x ) = 1 $ for all negative $ x $, since for $ x < 0 $ there is a negative $ y $, namely $ y = \frac { x + 2 - \sqrt { x ^ 2 + 4 } } 2 $, such that $ \frac { y ^ 2 - 2 y } { y - 1 } = x $. By \eqref{2}, that is also true for positive $ x $, and $ g $ is the constant one function, the other mentioned solution.
