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So what i am given is.

Ex 1.2.1

$x^{'} = x^{-1} $ and $x(0)=0$

we had a lengthy discuss that this was not discontinuous it simply didn't make any sense which i actually understood (or at least thought i did)

As practice we were asked to find an IVP with $x(0)=x_{0} $ for which the equation above is defined at $x_{0}$ but the solution does not exist. i know that the equation is deifned everywhere else but im still not sure what this is asking.

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    What's the question? – 2017-01-11
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    $(x')^2+x^2=-2,\,x(0)=0$ – 2017-01-11
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    sorry i mean the an $x_{0}$ for the given differential equation – 2017-01-11
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    http://www.wolframalpha.com/input/?i=y%27+%3D+y,+y(0)+%3D%3D+0 – 2017-01-11
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    You punched that in slightly wrong, http://www.wolframalpha.com/input/?i=y%27+%3D+1%2Fy,+y(0)+%3D%3D+0 is the correct answer if we're gonna put one – 2017-01-11

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You can work out this problem by first doing the general case then plug in $x(0) = 0$ at the end and the solution works out fine.

If your issue with this is to do with it's not being discontinuous, for this case the answer is as simple as $x'$ is not defined at 0 so it does not make sense to talk about continuity/discontinuity there.

The fact that $x'$ is not defined at $0$ is irrelevant, it still satisfies all of the properties given where both $x$ and $x'$ are defined (namely $x′=x^{−1}$ and $ x(0)=0$)