Find the partial derivative of f(x,y) with respect to y.

The given function is $f(x,y) = e^{2y} \cos(xy)$.
We use the product rule $\frac{\partial}{\partial y}(uv) = \frac{\partial u}{\partial y}v + u\frac{\partial v}{\partial y}$.
Let $u = e^{2y}$ and $v = \cos(xy)$.

First, find $\frac{\partial u}{\partial y}$:

$$\frac{\partial}{\partial y}(e^{2y}) = e^{2y} \cdot \frac{\partial}{\partial y}(2y) = 2e^{2y}$$

Next, find $\frac{\partial v}{\partial y}$:

$$\frac{\partial}{\partial y}(\cos(xy)) = -\sin(xy) \cdot \frac{\partial}{\partial y}(xy) = -\sin(xy) \cdot x = -x\sin(xy)$$

Now, apply the product rule:

$$f_y(x,y) = (2e^{2y})\cos(xy) + e^{2y}(-x\sin(xy))$$ $$f_y(x,y) = 2e^{2y}\cos(xy) - xe^{2y}\sin(xy)$$

Factor out $e^{2y}$:

$$f_y(x,y) = e^{2y}(2\cos(xy) - x\sin(xy))$$

Substitute $x=0$ and $y=\pi$ into the expression for $f_y(x,y)$:

$$f_y(0, \pi) = e^{2(\pi)}(2\cos(0 \cdot \pi) - 0 \cdot \sin(0 \cdot \pi))$$ $$f_y(0, \pi) = e^{2\pi}(2\cos(0) - 0 \cdot \sin(0))$$

Since $\cos(0) = 1$ and $\sin(0) = 0$:

$$f_y(0, \pi) = e^{2\pi}(2(1) - 0(0))$$ $$f_y(0, \pi) = e^{2\pi}(2 - 0)$$ $$f_y(0, \pi) = 2e^{2\pi}$$

The final answer is $2e^{2\pi}$.