Sure, I can help you solve this math problem step by step.

Let's begin by defining the given inequality and rearranging it to isolate one of the variables.

xf(y) + yf(x) <= 2

xf(y) <= 2 - yf(x)

f(y) <= (2/x) - (y/x)f(x)

Now, we have an expression for f(y) in terms of f(x).

Let's consider a fixed value of x and define a new function g(x) = 2/x. This function is continuous and decreasing for x > 0. Therefore, for any fixed value of x, there exists a unique value of y such that f(y) = g(x) - (y/x)f(x).

Let's plug this expression for f(y) back into the original inequality:

xf(y) + yf(x) <= 2

x(g(x) - (y/x)f(x)) + yf(x) <= 2

g(x) <= (y/x)f(x) + (2-yf(x))/x

Since g(x) is fixed for a given value of x, there exists a unique value of y that satisfies this inequality. Therefore, we can say that f(x) is a monotonically decreasing function.

Now, let's consider the case when x = y. In this case, the inequality becomes:

xf(x) + xf(x) <= 2

2xf(x) <= 2

f(x) <= 1

Since f(x) is monotonically decreasing, this implies that f(x) <= 1 for all x > 0.

Let's now consider the case when f(x) = 1 for some x. In this case, the inequality becomes:

xf(y) + y <= 2

y <= 2 - xf(y)

y/x <= 2/x - f(y)

Since y/x and 2/x are fixed, there exists a unique value of f(y) that satisfies this inequality. This implies that f(x) cannot be equal to 1 for any x > 0.

Therefore, the only function that satisfies the given conditions is f(x) = 1/x.