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More specifically :- I have to prove that $f(x)= 1-\sqrt{(1-x^2)}$ is a unique solution on [-1,1] for the DE: $y''=(1+(y')^2)^{3/2}, y(0)=1,y'(0)=1$

I am totally lost on this topic. So far, I tried to proceed as saying:- $g(y',(1+(y')^2)^{3/2})$ is a continuosly differentiable function and therefore Lipschitz on any compact set. and $f(x)$ is continuous and differentiable at 0. So maximize $||g(.)||_\infty $ in the neighborhood of (0,1) to find the radius in which this solution works using local Picard condition.

Am I even close to the right way to do this?

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    What are you allowed to assume? Do you know picard's theorem and are trying to prove that this function satisfies the hypotheses or are you starting from scratch? – 2017-01-18
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    @ I know the theorem and trying to prove it using that. – 2017-01-18

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Disclaimer: I have never done a problem like this and below is my take.

Note that a differentiable function is Lipschitz if and only if it has bounded derivative.

You then need to verify that the RHS of $y''=(1+(y')^2)^{3/2}=g(y')$ is Lipschitz. A change of variables, maybe taking $f=y'$ should convince you that modulo needing more initial conditions, this set up is exactly what Picard's theorem allows you to tackle.

So differentiating, we have $$ g'(y')=3(1+(y')^2)^{1/2}2y'\leq 6(1+(1+\delta))*(1+\delta)=6(2+\delta)(1+\delta) $$ and is thus locally Lipschitz around $y'(0)=1$, implying that a solution to the IVP is unique Verifying that what you have is in fact a solution involves differentiating a few times and plugging stuff in.