Question

Determine whether the graph of the equation is symmetric with respect to the y-axis, the x-axis, the origin, more than one of these, or none of these.

Answer 1

$\bf x^2y^2+3xy=1\\\\ -------------------------------\\\\ \stackrel{\stackrel{\textit{test for x-symmetry}}{y=-y}}{x^2(-y)^2+3x(-y)=1}\implies x^2y^2-3xy=1\impliedby \begin{array}{llll} \textit{function differs}\\ \textit{from original}\\ \textit{no dice} \end{array}\\\\ -------------------------------$

$\bf \stackrel{\stackrel{\textit{test for y-symmetry}}{x=-x}}{(-x)^2y^2+3(-x)y=1}\implies x^2y^2-3xy=1\impliedby \begin{array}{llll} \textit{function differs}\\ \textit{from original}\\ \textit{no dice} \end{array}\\\\ -------------------------------\\\\ \stackrel{\stackrel{\textit{test for origin-symmetry}}{x=-x~~y=-y}}{(-x)^2(-y)^2+3(-x)(-y)=1}\implies x^2y^2+3xy=1\impliedby \begin{array}{llll} origin\\symmetry \end{array}$

so, recall, the function has symmetry when the yielded resulting function resembles the original function, after negativizing the variable(s).

Also recall that minus*plus is minus, and minus*minus is plus.