$\left| {\begin{array}{cccccccccccccccccccc}{{y^2}{z^2}}&{yz}&{y + z}\\{{z^2}{x^2}}&{zx}&{z + x}\\{{x^2}{y^2}}&{xy}&{x + y}\end{array}} \right| = 0$

We have to prove,
$\left| {\begin{array}{cccccccccccccccccccc}{{y^2}{z^2}}&{yz}&{y + z}\\{{z^2}{x^2}}&{zx}&{z + x}\\{{x^2}{y^2}}&{xy}&{x + y}\end{array}} \right| = 0$

$\therefore$ ${\rm{LHS}} = \left| {\begin{array}{cccccccccccccccccccc}{{y^2}{z^2}}&{yz}&{y + z}\\{{z^2}{x^2}}&{zx}&{z + x}\\{{x^2}{y^2}}&{xy}&{x + y}\end{array}} \right|$

$= \frac{1}{{xyz}}\left| {\begin{array}{cccccccccccccccccccc}{x{y^2}{z^2}}&{xyz}&{xy + xz}\\{{x^2}y{z^2}}&{xyz}&{yz + xy}\\{{x^2}{y^2}z}&{xyz}&{xz + yz}\end{array}} \right|$

$= \frac{1}{{xyz}}{(xyz)^2}\left| {\begin{array}{cccccccccccccccccccc}{yz}&1&{xy + xz}\\{xz}&1&{yz + xy}\\{xy}&1&{xz + yz}\end{array}} \right|$

[taking $(xyz)$ common from ${C_1}$

and $\left. {{C_2}} \right]$
$= xyz\left| {\begin{array}{llllllllllllllllllll}{yz}&1&{xy + yz + zx}\\{xz}&1&{xy + yz + zx}\\{xy}&1&{xy + yz + zx}\end{array}} \right|\left[ {{C_3} \to {C_3} + {C_1}} \right]$

$= xyz(xy + yz + zx)\left| {\begin{array}{llllllllllllllllllll}{yz}&1&1\\{xz}&1&1\\{xy}&1&1\end{array}} \right|$

[taking $(xy + yz + zx)$ common from ${C_3}$]
$= 0$

[since,${C_2}$ and ${C_3}$ are identicals]
$= {\rm{RHS}}$