Function of multiple variables - Derivative problem Im trying to calculate the derivative $\frac{dz}{dt}$ with $z=e^{xy²}$ in $t=\frac{\pi}{2}$ with $x = tcos(t)$ and $y = tsin(t)$. 
I am however ending up with an answer that does not make much sense to me compared with the correct answer. This is my solution.
Applying the chainrule i get the following:
$$\frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$$
These are the derivatives that I find for this problem.
$$\frac{\partial z}{\partial x} = e^{xy²}*y² $$
$$\frac{dx}{dt} = cos(t)-sin(t)*t$$
$$\frac{\partial z}{\partial y} = e^{xy²}*2xy$$
$$\frac{dy}{dt} = cos(t)*t+sin(t)$$
After filling the result of my chainrule with the derivatives that I found, substituting all values for $t$ with $\frac{\pi}{2}$ and getting rid of the trig elements I end up with:
$$\frac{dz}{dt} = e^{xy²}*y²-(\frac{\pi}{2})+e^{xy²}*2xy$$
All this whilst the correct result should be $$-\frac{\pi³}{8}$$.
I don't see where I take a wrong turn. It'd be great if someone could point me in the right direction. Also please tell me if I chose the wrong name / title for this topic.

If $t=\frac{\pi}{2}$ then $x=0$ and $y=\frac{\pi}{2}$
so
$\frac{\partial z}{\partial x} = e^{xy²}*y²=\frac{\pi^2}{4}$
$\frac{dx}{dt} = cos(t)-sin(t)*t=-\frac{\pi}{2}$
$\frac{\partial z}{\partial y} = e^{xy²}*2xy=0$
then
$\frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}=-\frac{\pi^3}{8}$
