#y(t)=t/e^(3t)=te^(-3t)#
#:. y'(y)=t{d/dte^(-3t)}+e^(-3t)d/dt(t)#
#=t(e^(-3t))d/dt(-3t)+e^(-3t)(1)...[because," the Chain Rule]."#
#:. y'(t)=(1-3t)e^(-3t)..........(1).#
#x(t)=(t+3t^2)e^t#
#:. x"(t)=(t+3t^2)d/dt(e^t)+e^td/dt(t+3t^2)#
#=(t+3t^2)e^t+e^t(1+6t)#
#:. x'(t)=(t+3t^2+1+6t)e^t=(1+7t+3t^2)e^t............(2).#
Now, by the Rule for Parametric Diffn., we have,
#dy/dx=(y'(t))/(x'(t)), x'(t)!=0...........(star).#
Now, #x'(t)=0 rArr (1+7t+3t^2)e^t=0#
# rArr 1+7t+3t^2=0, because, e^t!=0, AA t in RR.#
# rArr t={-7+-sqrt(49-4(1)(3))}/2=(-7+-sqrt37)/2.#
# rArr t!=(-7+-sqrt37)/2, x'(t)!=0.#
#:., by (star), dy/dx={(1-3t)e^(-3t)}/{(1+7t+3t^2)e^t}, or, #
# dy/dx=(1-3t)/((1+7t+3t^2)e^(4t)), t!=(-7+-sqrt37)/2.#
Enjoy Maths.!