Find $dz/dt$ for $z=e^x \cos y$, where $x$ and $y$ are functions of $t$ defined by$$x^3+e^x-t^2-t=1\text{ and }yt^2+y^2t-t+y=0.$$
I try to use chain rule that dz/dt = (dz/dx)(dx/dt)+(dz/dy)(dy/dt), but I don't know how to find dx/dt and dy/dt in this problem.
For normal differentiation, $$\dfrac{dz}{dt}=-e^x\sin y\dfrac{dy}{dt}+\cos ye^x\dfrac{dx}{dt}$$ Implicit differentation thus gives $$3x^2\dfrac{dx}{dt}+e^x\dfrac{dx}{dt}-2t-1=0\\ t^2\dfrac{dy}{dt}+2ty+y^2+2ty\dfrac{dy}{dt}-1+\dfrac{dy}{dt}=0$$ Thus, $$3x^2\dfrac{dx}{dt}+e^x\dfrac{dx}{dt}-2t-1=0\\ \implies \dfrac{dx}{dt}=\dfrac{1+2t}{3x^2+e^x}$$ And $$t^2\dfrac{dy}{dt}+2ty+y^2+2ty\dfrac{dy}{dt}-1+\dfrac{dy}{dt}=0\\ \implies \dfrac{dy}{dt}=\dfrac{1-2ty+y^2}{t^2+2ty+1}$$ Substitute: $$\boxed{\dfrac{dz}{dt}=-e^x\sin y\dfrac{1-2ty+y^2}{t^2+2ty+1}+e^x\cos y\dfrac{1+2t}{3x^2+e^x}}$$