If we have an equation as \[{{x}^{2}}+{{y}^{2}}=1\] then,
1). \[yy''-2{{(y')}^{2}}+1=0\]
2). \[yy''+{{(y')}^{2}}+1=0\]
3). \[yy''-{{(y')}^{2}}-1=0\]
4). \[yy''+2{{(y')}^{2}}+1=0\]

Hint: First of all differentiate the given term with respect to \[x\] on both sides then take \[2\]common from the obtained differential to make it in simple form after that again differentiate the obtained differential with respect to \[x\] on both the sides to check which option is correct in the given options.

Complete step-by-step solution:
The process of finding the differential coefficient of a function is called differentiation or we can say that differentiation is a process where we find the instantaneous rate of change in function based on one of its variables.
The derivative or differential coefficient of a function can be obtained directly by the definition of differentiation without using addition, multiplication and quotient formulae of differentiation.
The coefficient of \[y\] with respect to \[x\] is represented by \[\dfrac{dy}{dx}\] .
We know that the meaning of \[dx\] is the increment in \[x\] and it may be positive or negative. Similarly \[dy\]means the increment in the value of \[y\]
If there is an increment in the value of \[x\] and \[y\] in the same direction either both positive or both negative then the value of \[\dfrac{dy}{dx}\] is always positive. On the other hand if the increments in \[x\] and \[y\] are of positive direction that is one positive and other negative then the value of \[\dfrac{dy}{dx}\] will be negative.
Let \[x\] be a variable quantity, after some increment its value \[{{x}_{1}}\] becomes \[{{x}_{2}}\] . the difference between \[{{x}_{1}}\] and \[{{x}_{2}}\] is called the change in value of \[x\]. It is represented by \[\delta x\]. The change \[\delta x\] may be positive or negative.
Let us suppose that \[y\] is a function of \[x\] and if \[\delta x\] be the change in \[x\] then the corresponding change in \[y\] be denoted by \[\delta y\].
Now according to the question:
We have given a term \[{{x}^{2}}+{{y}^{2}}=1\]
Differentiate the given term with respect to \[x\] on both sides
\[\Rightarrow \dfrac{d}{dx}({{x}^{2}})+\dfrac{d}{dx}({{y}^{2}})=\dfrac{d}{dx}(1)\]
\[\Rightarrow 2x+2y\dfrac{dy}{dx}=0\]
We know that we can write differential of \[y\] as \[y'\]
\[\Rightarrow 2x+2yy'\]
Now take \[2\] common from left hand side we will get:
\[\Rightarrow 2(x+yy')=0\]
\[\Rightarrow (x+yy')=0\]
Now again differentiate the obtained differential with respect to \[x\] on both sides:
\[\Rightarrow \dfrac{d}{dx}x+\dfrac{d}{dx}yy'=\dfrac{dy}{dx}(0)\]
Apply product rule in differentiation of \[yy'\] :
\[\Rightarrow 1+y'\cdot \dfrac{dy}{dx}+y\cdot \dfrac{d}{dx}(y')=0\]
Differentiation of \[y'=\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]
\[\Rightarrow 1+y'\cdot y'+y\cdot \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=0\]
You can write \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\] as \[y''\]
\[\Rightarrow 1+{{(y')}^{2}}+y\cdot y''=0\]
\[\Rightarrow y\cdot y''+{{(y')}^{2}}+1=0\]
Hence we can say that option \[(2)\] is correct.

Note: Students must understand the difference between \[\dfrac{\delta y}{\delta x}\] and \[\dfrac{dy}{dx}\] . Here \[\dfrac{\delta y}{\delta x}\] is a fraction with \[\delta y\] as a numerator and \[\delta x\] as a denominator while \[\dfrac{dy}{dx}\] is not a fraction. It is a limiting value of \[\dfrac{\delta y}{\delta x}\] . The differential coefficients of those functions which start with \['co'\] like \[\text{cosx,cotx,cosecx}\] are always negative.