Understanding the Problem: Partial Derivatives and Euler's Theorem

The problem asks for the value of the expression \(x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y}\) for the given function \(U = \mathop e\nolimits^{\frac{{\mathop x\nolimits^2 }}{{\mathop y\nolimits^2 }}} + \mathop e\nolimits^{\frac{{\mathop y\nolimits^2 }}{{\mathop x\nolimits^2 }}} \). This specific form of expression \(x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y}\) is strongly related to Euler's theorem on homogeneous functions.

Checking for Homogeneity

First, let's determine if the function \(U(x, y)\) is a homogeneous function. A function \(f(x, y)\) is said to be homogeneous of degree \(n\) if for any non-zero constant \(t\), \(f(tx, ty) = t^n f(x, y)\).

Let's substitute \(tx\) for \(x\) and \(ty\) for \(y\) in the function \(U\):

\(U(tx, ty) = \mathop e\nolimits^{\frac{{\mathop {(tx)}\nolimits^2 }}{{\mathop {(ty)}\nolimits^2 }}} + \mathop e\nolimits^{\frac{{\mathop {(ty)}\nolimits^2 }}{{\mathop {(tx)}\nolimits^2 }}} \)

\(U(tx, ty) = \mathop e\nolimits^{\frac{{\mathop t\nolimits^2 \mathop x\nolimits^2 }}{{\mathop t\nolimits^2 \mathop y\nolimits^2 }}} + \mathop e\nolimits^{\frac{{\mathop t\nolimits^2 \mathop y\nolimits^2 }}{{\mathop t\nolimits^2 \mathop x\nolimits^2 }}} \)

Since \(t \neq 0\), we can cancel out \(\mathop t\nolimits^2\) from the exponents:

\(U(tx, ty) = \mathop e\nolimits^{\frac{{\mathop x\nolimits^2 }}{{\mathop y\nolimits^2 }}} + \mathop e\nolimits^{\frac{{\mathop y\nolimits^2 }}{{\mathop x\nolimits^2 }}} \)

We observe that \(U(tx, ty) = U(x, y)\). This can be written as \(U(tx, ty) = t^0 U(x, y)\).

Thus, the function \(U\) is a homogeneous function of degree \(n = 0\).

Applying Euler's Theorem

Euler's theorem for homogeneous functions states that if \(U(x, y)\) is a homogeneous function of degree \(n\), then the following relationship holds:

\(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = nU\)

In our case, we found that \(U\) is a homogeneous function of degree \(n = 0\).

Substituting \(n=0\) into Euler's theorem, we get:

\(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = 0 \cdot U\)

\(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = 0\)

Conclusion

Based on Euler's theorem, for the given homogeneous function \(U\) of degree zero, the expression \(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}}\) evaluates to 0.

The final answer is \textbf{0}.