3. The number of all real number pairs $(x, y)$ that satisfy the system of equations
$$
\left\{\begin{array}{l}
x^{x+y}=y^{x-y} \\
x^{3} y=1
\end{array},\right.
$$
is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
3. D.

When $x=1$ or -1, from equation (2) we get $y= \pm 1$. Therefore, $(x, y)=(1,1),(-1,-1)$, both satisfy equation (1). When $x \neq \pm 1$, from equation (2) we get
$$
y=x^{-3} \text {. }
$$

Substituting equation (3) into equation (1) gives
$$
x^{x+y}=x^{3 y-3 x} \Rightarrow x+y=3 y-3 x \Rightarrow y=2 x \text {. }
$$

Substituting equation (4) into equation (2) gives
$$
x^{4}=\frac{1}{2} \Rightarrow x= \pm \frac{1}{\sqrt[4]{2}} \text {. }
$$

Therefore, $y= \pm \sqrt[4]{8}$.
Hence, $(x, y)=\left(\frac{1}{\sqrt[4]{2}}, \sqrt[4]{8}\right),\left(-\frac{1}{\sqrt[4]{2}},-\sqrt[4]{8}\right)$.