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ĐK: $\left\{ \begin{array}{l}
x > 0,\,\,y > 0\\
x \ne 1,\,\,y \ne 1
\end{array} \right.$
$\begin{array}{l}
(1)\,\, \Leftrightarrow \,\,\left\{ \begin{array}{l}
3x + 2y = {x^2}\\
2x + 3y = {y^2}
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\left\{ \begin{array}{l}
3x + 2y = {x^2}\\
x - y = {x^2} - {y^2}
\end{array} \right.\\
\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\left\{ \begin{array}{l}
3x + 2y = {x^2}\\
\left( {x - y} \right)\left( {x + y - 1} \right) = 0\,\,\,
\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\left[ \begin{array}{l}
\left\{ \begin{array}{l}
3x + 2y = {x^2}\\
x - y = 0
\end{array} \right.\,\,\,\,\,\,\,\,\,(2)\\
\left\{ \begin{array}{l}
3x + 2y = {x^2}\\
x + y - 1 = 0
\end{array} \right.\,\,\,\,\,\,\,\,\,(3)
\end{array} \right.
\end{array}$
$\begin{array}{l}
(2)\,\, \Leftrightarrow \,\left\{ \begin{array}{l}
x = y\\
5x = {x^2}
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\left\{ \begin{array}{l}
x = y = 0\,\,\,\,\,\,(l)\\
x = y = 5
\end{array} \right.\\
\\
(3)\,\, \Leftrightarrow \left\{ \begin{array}{l}
y = 1 - x\\
3x + 2\left( {1 - x} \right) = {x^2}
\end{array} \right.\,\,\,\,\, \Leftrightarrow \,\,\,\left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = - 1\\
y = 2
\end{array} \right.\,\,\,\,\,\,\,\,\,(loại)\\
\left\{ \begin{array}{l}
x = 2\\
y = - 1
\end{array} \right.\,\,\,\,\,\,\,\,\,(loại)
\end{array} \right.
\end{array}$
Vậy hệ phương trình có $1$ nghiệm duy nhất là $(5;\,5)$
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