3. In the Cartesian coordinate system, the set of points $(x, y)$ that satisfy $x<y,|x|<3,|y|<3$, and make the equation $\left(x^{3}-y^{3}\right) t^{4}+$ $(3 x+y) t^{2}+\frac{1}{x-y}=0$ have no real roots is denoted as $N$. Then the area of the region formed by the point set $N$ is $\qquad$
3. $\frac{81}{5}$ Detailed Explanation: Let $u=t^{2}$, the original equation becomes $\left(x^{3}-y^{3}\right) u^{2}+(3 x+y) u+\frac{1}{x-y}=0$.
$$
\Delta=(3 x+y)^{2}-4\left(x^{3}-y^{3}\right) \cdot \frac{1}{x-y}=5 x^{2}+2 x y-3 y^{2}=(5 x-3 y)(x+y) \text {. }
$$

The given equation has no real roots if and only if equation (1) has no real roots or has real roots but all are negative, so,
$$
\left\{\begin{array} { l } 
{ x < y , } \\
{ | x | < 3 , } \\
{ | y | < 3 , } \\
{ ( 5 x - 3 y ) ( x + y ) < 0 , }
\end{array} \text { or } \left\{\begin{array}{l}
x<y, \\
|x|<3, \\
|y|<3, \\
(5 x-3 y)(x+y) \geqslant 0, \\
3 x+y<0 .
\end{array}\right.\right.
$$

The area of the region formed by the point set $N$ is $\frac{1}{2} \times \frac{24}{5} \times 3+\frac{1}{2} \times 6 \times 3=\frac{81}{5}$.