Let $$f : \mathbb{R} \to \mathbb{R}$$ be a function defined by $$f(x) = \max\{x, x^2\}$$. Let $$S$$ denote the set of all points in $$\mathbb{R}$$, where $$f$$ is not differentiable. Then:

We are given the real-valued function $$f:\mathbb R\to\mathbb R$$ defined by $$f(x)=\max\{x,\;x^2\}\,.$$

To analyse differentiability we first express $$f$$ as a piece-wise function, because at each real number we must decide which of the two expressions, $$x$$ or $$x^2$$, is larger.

We compare $$x^2$$ with $$x$$:

$$x^2\ge x\;\Longleftrightarrow\;x^2-x\ge0\;\Longleftrightarrow\;x(x-1)\ge0.$$

The inequality $$x(x-1)\ge0$$ holds when $$x\le0$$ or $$x\ge1$$, and it fails when $$0<x<1$$. Hence

$$ f(x)= \begin{cases} x^2,& x\le0,\\[4pt] x, & 0\lt x\lt 1,\\[4pt] x^2,& x\ge1. \end{cases} $$

Now we differentiate each branch separately.

For $$x<0$$ we have $$f(x)=x^2,$$ so using the standard rule $$\dfrac{d}{dx}(x^n)=nx^{n-1}$$ we get

$$f'(x)=2x\qquad(x<0).$$

For $$0<x<1$$ we have $$f(x)=x,$$ and the derivative of $$x$$ is

$$f'(x)=1\qquad(0<x<1).$$

For $$x>1$$ we have again $$f(x)=x^2,$$ giving

$$f'(x)=2x\qquad(x>1).$$

Thus $$f'$$ is well defined everywhere except possibly at the junction points $$x=0$$ and $$x=1$$. We must examine these two points separately by comparing the left-hand and right-hand derivatives.

Point $$x=0$$.

The left-hand derivative is the limit of $$2x$$ as $$x\to0^-$$:

$$\lim_{x\to0^-}2x=0.$$

The right-hand derivative is the limit of $$1$$ as $$x\to0^+$$:

$$\lim_{x\to0^+}1=1.$$

Because $$0\ne1$$, the two one-sided derivatives are unequal, so $$f$$ is not differentiable at $$x=0$$.

Point $$x=1$$.

The left-hand derivative is the limit of $$1$$ as $$x\to1^-$$:

$$\lim_{x\to1^-}1=1.$$

The right-hand derivative is the limit of $$2x$$ as $$x\to1^+$$:

$$\lim_{x\to1^+}2x=2.$$

Since $$1\ne2$$, the one-sided derivatives differ, so $$f$$ is not differentiable at $$x=1$$.

For every other real number the function matches just one smooth branch, and the derivative we calculated is finite and continuous, hence $$f$$ is differentiable everywhere except at $$0$$ and $$1$$.

Therefore the set $$S$$ of non-differentiability points is

$$S=\{0,1\}.$$

Among the given options, this set corresponds to Option A.

Hence, the correct answer is Option A.