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We can solve the simultaneous equations
\[x^3y_2=72\]
  (1)
\[x^2y^4=324\]
  (2)
May making the powers of either  
\[x\]
  or  
\[y\]
  the same, then dividing one equation by the other, which cancels either  
\[x\]
  or  
\[y\]
.
We can make the coefficients of  
\[y\]
  the same by squaring equation )1_. We obtain
\[x^6y_4=72\]
  (3)
(3) divided by (2) gives
\[\frac{x^6y_4}{x^2y^4}=\frac{72^2}{324} x^4=15 \rightarrow x=-2, \: 2\]

In fact  
\[x\]
  cannot be equal to -2 because from the first equation
\[x^3 = \frac{72}{y^2} \gt 0\]
.
Then from (1),
\[y=\sqrt{\frac{72}{x^3}}=\sqrt{\frac{72}{2^3}}=-3, \: 3\]
.
\[x=2, \: y=-3\]
  and
\[x=2, \: y=3\]
  satisfy both equations so are the solutions.