Find the product of all real values of $y$ such that  $|{-2y}|\cdot\left|{-\dfrac{y}{2}}\right| = 36$.
To solve for $y$, I need to get rid of the absolute value signs. I recall that $|x| = x$ if $x \geq 0$, and $|x| = -x$ if $x < 0$. So, depending on the sign of $-2y$ and $-\dfrac{y}{2}$, I have four possible cases to consider. Case 1: $-2y \geq 0$ and $-\dfrac{y}{2} \geq 0$. This means that $y \leq 0$ and $y \leq 0$, so $y \leq 0$. In this case, $|{-2y}|\cdot\left|{-\dfrac{y}{2}}\right| = (-2y)(-\dfrac{y}{2}) = y^2$. Setting this equal to 36, I get $y^2 = 36$. Taking the square root of both sides, I get $y = \pm 6$. However, since $y \leq 0$, I only accept the negative value, $y = -6$. Case 2: $-2y \geq 0$ and $-\dfrac{y}{2} < 0$. This means that $y \leq 0$ and $y > 0$, which is impossible. So there are no solutions in this case. Case 3: $-2y < 0$ and $-\dfrac{y}{2} \geq 0$. This means that $y > 0$ and $y \leq 0$, which is also impossible. So there are no solutions in this case. Case 4: $-2y < 0$ and $-\dfrac{y}{2} < 0$. This means that $y > 0$ and $y > 0$, so $y > 0$. In this case, $|{-2y}|\cdot\left|{-\dfrac{y}{2}}\right| = (-2y)(-\dfrac{y}{2}) = y^2$. Setting this equal to 36, I get $y^2 = 36$. Taking the square root of both sides, I get $y = \pm 6$. However, since $y > 0$, I only accept the positive value, $y = 6$. Therefore, the only real values of $y$ that satisfy the equation are $y = -6$ and $y = 6$. The product of these values is $(-6)(6) = -36$.

