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In order to prove this statement true or false, I need to find the contra-positive of the equation below. I have already tried to prove this by negation which was not successful.

$\ \exists \space x \in \Bbb R \space, \exists \space y \in \Bbb R \space| \space \frac{1}{x}+\frac{1}{y}=\frac{1}{x+y} $

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    Don't you mean $\frac{1}{x}+\frac{1}{y}=\frac{1}{x+y}$? – 2017-01-24

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Contrapositives are about conditionals ('if .. then' statements), which you don't have here!

There is no such thing as the contrapositive of an identity (equality) or equation.

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    Then how would I go about proving that this is true or false? – 2017-01-24
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    @Dain Well, the claim seems unlikely, doesn't it? So: we suspect the claim is false. To prove that, you really need to do some algebra, e.g. go from $\frac{1}{x}+\frac{1}{y}=\frac{1}{x+y} $ to $\frac{y}{x*y}+\frac{x}{x*y}=\frac{1}{x+y} $ to $\frac{x+y}{x*y}=\frac{1}{x+y} $ to $(x + y)^2 = x*y$ to $x^2 + 2*x*y + y^2 = x*y$ to $x^2 + x*y + y^2 = 0$ ... and now you see the problem: this only works for $x$ and $y$ being $0$ ... but that would mean we'd be dividing by $0$ in the original equation. So this is just not going to work! – 2017-01-24