The function $$f(x) = |x^2 - 2x - 3| \cdot e^{9x^2-12x+4}$$ is not differentiable at exactly:

We have a composite function

$$f(x)=\;|x^{2}-2x-3|\; \cdot e^{\,9x^{2}-12x+4}$$

To locate the points where $$f(x)$$ is not differentiable, we first note that the exponential part $$e^{\,9x^{2}-12x+4}$$ is an elementary exponential function. An exponential function of the form $$e^{g(x)}$$ is differentiable for every real number because its derivative is

$$\dfrac{d}{dx}\bigl(e^{g(x)}\bigr)=g'(x)\,e^{g(x)},$$

and both $$g(x)$$ and $$g'(x)$$ are polynomials here. Therefore, any loss of differentiability can occur only because of the modulus (absolute-value) factor $$|x^{2}-2x-3|$$.

A function of the type $$|u(x)|$$ is differentiable everywhere except possibly at those values of $$x$$ where its inside expression $$u(x)$$ equals zero. Hence, we first solve

$$x^{2}-2x-3 = 0.$$

We factor the quadratic:

$$x^{2}-2x-3=(x-3)(x+1).$$

Setting each factor to zero gives

$$x-3=0 \;\Longrightarrow\; x = 3,$$

$$x+1=0 \;\Longrightarrow\; x = -1.$$

Thus the inside of the modulus vanishes precisely at

$$x=-1 \quad\text{and}\quad x=3.$$

Next, we verify that differentiability indeed fails at these two points. Whenever $$u(x)=0,$$ the expression $$|u(x)|$$ changes from $$u(x)$$ on one side to $$-u(x)$$ on the other. This causes the left-hand derivative (LHD) and the right-hand derivative (RHD) of the overall function to differ, because the sign of $$u(x)$$ flips. Consequently the derivative does not exist at those points.

At every other point, $$x^{2}-2x-3\neq0,$$ so the modulus simply contributes a factor of either $$+1$$ or $$-1,$$ and both the modulus and the exponential part are smooth there. Hence the product remains differentiable.

Therefore the function loses differentiability only at

$$x=-1 \quad\text{and}\quad x=3,$$

giving exactly two distinct points of non-differentiability.

Hence, the correct answer is Option B.