Answer
$$\frac{dz}{dt} = \frac {x/t -ysin \space t} { \sqrt {1 + x^2 + y^2}} $$
Work Step by Step
According to the Chain Rule:
$$\frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} +\frac{\partial z}{\partial y} \frac{dy}{dt} $$
$$ \frac{\partial z}{\partial x} = \frac{\partial (\sqrt{1 + x^2 + y^2})}{\partial x} = \frac{1}{2 \sqrt {1 + x^2 + y^2}} (2x) = \frac{x}{ \sqrt {1 + x^2 + y^2}}$$ $$\frac{dx}{dt} = \frac{d(ln \space t)}{dt} = 1/t$$ $$\frac{\partial z}{\partial y} = \frac{\partial ( \sqrt {1 + x^2 + y^2})}{\partial y} = \frac{1}{2 \sqrt {1 + x^2 + y^2}}(2y) = \frac y { \sqrt {1 + x^2 + y^2}} $$ $$ \frac{dy}{dt} = \frac{d(cos \space t)}{dt} = -sin \space t$$
Therefore:
$$\frac{dz}{dt} =\Bigg( \frac x { \sqrt {1 + x^2 + y^2}} \Bigg) (1/t)+ \Bigg( \frac y { \sqrt {1 + x^2 + y^2}} \Bigg)(-sin \space t)$$$$\frac{dz}{dt} = \frac {x/t -ysin \space t} { \sqrt {1 + x^2 + y^2}} $$
