Answer
$-\frac{7}{2}$
Work Step by Step
Since $x=-1$, $y=2$, $z=-3$
We take into account the order of operations:
Parenthesis, Exponents, Multiplication/Division, Addition/Subtraction
Thus, we have:
$$\frac{(3x^2+2)^2-y^2}{6-3x^2y^2}=\frac{(3(-1)^2+2)^2-2^2}{6-3(-1)^22^2}\\
=\frac{(3+2)^2-4}{6-3(4)}=\frac{5^2-4}{6-12}=\frac{21}{-6}=-\frac{7}{2}
.$$
