91Ó°ÊÓ

First, let's analyze the given function: $$f(x, y)=\frac{x y}{x^{2} y^{2}+1}$$ Notice that the denominator in this expression, \(x^2y^2+1\), is always positive for all values of \(x\) and \(y\). This is because the product of two square numbers is always non-negative and adding \(1\) to a non-negative number always results in a positive number. So, the denominator can never be zero, which implies that there are no singularities (values of \(x\) and \(y\) for which the function is not defined) in this function.

Recall that a function is continuous at a point \((a, b)\) if the following conditions are met: 1. The function is defined at \((a, b)\). 2. The limit of the function, as \((x, y)\) approaches \((a, b)\), equals its value at \((a, b)\). Since our function is always defined (as noted in Step 1), it meets the first condition for all points in \(\mathbb{R}^2\). Next, we need to determine if the limit of the function as \((x, y)\) approaches \((a, b)\) exists for all points \((a, b)\). To do this, we can express the function as a polar coordinate function: $$f(r, \theta) = \frac{r^2 \sin \theta\cos \theta}{(r^4 \sin^2\theta\cos^2\theta) + 1}$$ This transformation makes it easier to work with and analyze the function. Observe that as \(r\) approaches \(0\), the numerator, \(r^2 \sin \theta\cos \theta\), approaches \(0\) as well. Moreover, the denominator, \((r^4 \sin^2\theta\cos^2\theta) + 1\), approaches \(1\). This means the limit of the function as \(r\) approaches \(0\) is equal to: $$\lim_{r \to 0} \frac{r^2 \sin \theta\cos \theta}{(r^4 \sin^2\theta\cos^2\theta) + 1} = \frac{0}{1} = 0$$ Since the limit of the function with respect to \((a,b)\) approaches \(0\), it meets the second condition for continuity.

Based on our analysis, the function \(f(x, y)\) meets all the conditions for continuity at all points in \(\mathbb{R}^2\). This is because the denominator can never be zero, and the limit of the function as \((x, y)\) approaches any point \((a, b)\) always exists. Therefore, the function \(f(x, y)\) is continuous at all points in \(\mathbb{R}^2\).