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This is what I did to prove that:

proof that it's an odd function

But I am not sure if the proof is valid.

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    Within the calculation, you assumed what you want to prove. So what you really proved in the photo is a tautological statement "if $f$ is odd, then $f$ is indeed odd!". – 2017-02-16
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    Here is one possible valid approach: Begin the proof by writing$$ f\left(-\frac{x+y}{x-y}\right) = f\left(\frac{y+x}{y-x}\right). $$ – 2017-02-16
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    Too confuse, can't tell if it is valid. – 2017-02-16

2 Answers 2

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Well, even if $f (-\alpha ) = f (\alpha) $, then $$\frac {f (-y)+f (-x)}{f (-y)-f (-x)} = \frac {f (y)+f (x)}{f (y)-f (x)} $$ $$=-\frac {f (x)+f (y)}{f (x)-f (y)} $$ $$=-f (\frac {x+y}{x-y}) $$

Hope it helps.

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If you swap the roles of the variables $x$ and $y$, the argument in the LHS gets negated and the value of the RHS gets negated as well.

$$f\left(\frac{y+x}{y-x}\right)=f\left(-\frac{x+y}{x-y}\right),$$

$$\frac{f(y)+f(x)}{f(y)-f(x)}=-\frac{f(x)+f(y)}{f(x)-f(y)}.$$

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    I am a bit embarrassed now that I didn't see this previously. Thanks for the answer. – 2017-02-16
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    Accepting my answer might disembarrass you :) – 2017-02-16