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The equation given is:

\[
f(x)f(y) = f(x) + f(y) + \frac{1}{f(x)f(y)}
\]

We want to find the function \( f(x) \).

### Step 1: Simplify the equation

First, let's rearrange the equation by multiplying both sides by \( f(x)f(y) \):

\[
f(x)^2 f(y)^2 = f(x)f(y) \left(f(x) + f(y)\right) + 1
\]

Now, let's expand the right-hand side:

\[
f(x)^2 f(y)^2 = f(x)^2 f(y) + f(x) f(y)^2 + 1
\]

### Step 2: Analyze possible solutions

Given the structure of the equation, a potential solution is to try simple functions like constant functions or linear functions. Let's try a linear function \( f(x) = x + c \) where \( c \) is a constant.

### Testing \( f(x) = 1 \):

Let \( f(x) = 1 \) for all \( x \):

\[
1 \times 1 = 1 + 1 + \frac{1}{1 \times 1}
\]

This simplifies to:

\[
1 = 2 + 1
\]

which is false, so \( f(x) = 1 \) is not a solution.

### Testing \( f(x) = x \):

Let \( f(x) = x \):

\[
xy = x + y + \frac{1}{xy}
\]

Multiplying both sides by \( xy \) gives:

\[
x^2 y^2 = xy(x + y) + 1
\]

Expanding this:

\[
x^2 y^2 = x^2 y + xy^2 + 1
\]

This simplifies to:

\[
x^2 y^2 - x^2 y - xy^2 = 1
\]

Which is true only in specific cases or with a particular domain.

### Testing \( f(x) = \frac{1}{x} \):

Let's consider \( f(x) = \frac{1}{x} \):

\[
\frac{1}{x} \times \frac{1}{y} = \frac{1}{x} + \frac{1}{y} + \frac{1}{\frac{1}{x} \times \frac{1}{y}}
\]

Simplifying:

\[
\frac{1}{xy} = \frac{1}{x} + \frac{1}{y} + xy
\]

This equation generally does not hold either.

### Testing \( f(x) = \frac{1}{\sqrt{x}} \):

This doesn't seem to yield simple results either. Given that typical simple forms of functions do not satisfy this equation, it suggests the solution might involve a more complex or non-standard function. Further testing of more sophisticated or piecewise functions might be required.

If you would like further steps or have specific conditions you'd like to consider, let me know! 

### Further Exploration Questions:

1. What happens if we substitute \( f(x) = ax + b \) into the equation?
2. Can we determine \( f(x) \) if we restrict the domain of \( x \) and \( y \)?
3. What if \( f(x) \) is assumed to be a trigonometric function?
4. What if we try \( f(x) = e^{g(x)} \) for some function \( g(x) \)?
5. Is there a piecewise function that can satisfy the equation?

### Tip:
When solving functional equations, testing simple functions like constants, linear, or power functions can often reveal patterns or help guide toward the solution.

Explore the solution of a functional equation involving f(x) and f(y). Learn how to solve complex functional equations and analyze possible solutions step-by-step. Suitable for undergraduate mathematics students.

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