It's multiplication btn 2y 2xy

The original equation is $x^2 \sec x - xy + 2y^2 - 2xy = 15$.
This simplifies to $x^2 \sec x - 3xy + 2y^2 = 15$.

When we differentiate implicitly with respect to $x$:

1. For the term $-3xy$: We use the product rule $\frac{d}{dx}(uv) = u'v + uv'$. Let $u = -3x$ and $v = y$. Then $u' = -3$ and $v' = \frac{dy}{dx}$. So, $\frac{d}{dx}(-3xy) = (-3)(y) + (-3x)\left(\frac{dy}{dx}\right) = -3y - 3x \frac{dy}{dx}$.

2. For the term $2y^2$: We use the chain rule $\frac{d}{dx}(f(y)^n) = n f(y)^{n-1} \frac{df}{dx}$. Here, $f(y) = y$ and $n=2$. So, $\frac{d}{dx}(2y^2) = 2 \cdot (2y^{2-1}) \cdot \frac{dy}{dx} = 4y \frac{dy}{dx}$.

Combining these, the derivative of $-3xy + 2y^2$ is indeed $-3y - 3x \frac{dy}{dx} + 4y \frac{dy}{dx}$.

The differentiation in the original solution was correct.