6. Find the integer solutions of the equation $x \sqrt{14-3 y^{2}}-y \sqrt{21-3 x^{2}}=7 \sqrt{2}$.
6. According to the Cauchy-Schwarz inequality,
\[
\left(x \sqrt{14-3 y^{2}}-y \sqrt{21-3 x^{2}}\right)^{2} \leqslant\left[x^{2}+\left(\sqrt{7-x^{2}}\right)^{2}\right] \cdot\left[\left(\sqrt{14-3 y^{2}}\right)^{2}+(-\sqrt{3} y)^{2}\right],
\]
since \( x \sqrt{14-3 y^{2}}-y \sqrt{21-3 x^{2}}=7 \sqrt{2} \),
then
\[
\left(x \sqrt{14-3 y^{2}}-y \sqrt{21-3 x^{2}}\right)^{2}=\left[x^{2}+\left(\sqrt{7-x^{2}}\right)^{2}\right] \cdot\left[\left(\sqrt{14-3 y^{2}}\right)^{2}+(-\sqrt{3} y)^{2}\right]=98.
\]
According to the condition for equality in the Cauchy-Schwarz inequality, we have and only have
\[
\left\{\begin{array}{l}
x=k \sqrt{14-3 y^{2}}, \\
\sqrt{7-x^{2}}=k(-\sqrt{3} y) .(k \text { is a constant })
\end{array}\right. \text {. }
\]

Eliminating \( k \), simplifying and rearranging, we get
\[
2 x^{2}+3 y^{2}=14 \text { and } x \text { and } y \text { have opposite signs } (x y<0) \text {. }
\]

From the original equation, we know \( -\sqrt{7} \leqslant x \leqslant \sqrt{7} \),
and \( -\frac{\sqrt{42}}{3} \leqslant y \leqslant \frac{\sqrt{42}}{3} \),
solving the indeterminate equation \( 2 x^{2}+3 y^{2}=14 \),
we get the integer solutions of the original equation as
\[
\left\{\begin{array}{l}
x=1, \\
y=-2 .
\end{array}\right.
\]