Answer
$-1.55$
Work Step by Step
When $t=1$, then $x_t=-sin(t)=-sin1$
and $y=sin(t)=sin1 \implies \dfrac{dy}{dt}=cos(t)=cos \ 1$
$z=cos(2t)=cos2$ when $t=1$
Next, $\dfrac{dz}{dt}=-sin(2t) \implies z_{t}(t=1)=- \sin 2$
Now, $f(x,y,z)=xy+yz+xz$
$f_{x}=y+z=sin \ 1+cos \ 2$( when $t=1$)
$f_{y}=x+z=cos1+cos2$ (when $t=1$)
$f_{z}=x+y=cos 1+sin \ 1$( when $t=1$)
Thus, $\dfrac{df}{dt}=f_{x}\dfrac{dx}{dt}+f_{y}\dfrac{dy}{dt}+f_{z}\dfrac{dz}{dt} \\=(\sin \ 1+\cos \ 2)(-sin \ 1)+(\cos \ 1+\cos \ 2)(\cos \ 1)+(\cos \ 1+\sin \ 1)(-\sin \ 2)=-1.55$
