To determine the points of discontinuity of the function \( f(x, y) \), we need to analyze where \( f \) is not continuous. The function is defined piecewise:
\(f(x, y) = \begin{cases} (x^2 - 1)^2 \cos^2\left(\frac{y^2}{(x^2 - 1)^2}\right) & \text{if } x \neq \pm 1 \\ 0 & \text{if } x = \pm 1\end{cases}\)
For \(x \neq \pm 1\), \( f(x, y) \) is clearly defined and continuous since it is a composition of continuous functions (polynomial and cosine functions), provided the denominator \( (x^2 - 1)^2 \) does not cause a problem when approaching 0.
We need to investigate what happens as \( x \to \pm 1 \).
Consider the limit:
\(\lim_{x \to 1} (x^2 - 1)^2 \cos^2\left(\frac{y^2}{(x^2 - 1)^2}\right)\)
As \(x \to 1\), \( (x^2 - 1)^2 \to 0 \). The problematic term is \( \cos^2\left( \frac{y^2}{(x^2 - 1)^2} \right) \), which oscillates rapidly because \( \frac{y^2}{(x^2 - 1)^2} \to \infty \), causing the limit to be indeterminate due to oscillation in the cosine term.
If we analyze similarly for \( x \to -1 \), the same issue arises.
Thus, \( f(x, y) \) is discontinuous at \( x = \pm 1 \) for every \( y \).
The discontinuity happens along the entire lines \( x = 1 \) and \( x = -1 \). These represent infinitely many points of discontinuity across the y-axis.
Since the function is discontinuous along two entire lines, the answer is implying infinity, which is beyond the expected solution range. Interpret this result as the conceptual description instead of numerically finite.
The expected number of points is not a numeric value within a finite range like 0,0, as discontinuity stretches across two lines. Thus, no simple count fulfills the point-based qualitative interpretation.