Answer
$\dfrac{ds}{dt} = \dfrac{x{\frac{dx}{dt}+y\frac{dy}{dt}+z\frac{dz}{dt}}}{\sqrt{(x^2+y^2+z^2)}}$
$\dfrac{ds}{dt} = \dfrac{(y\frac{dy}{dt}+z\frac{dz}{dt})}{\sqrt{(x^2+y^2+z^2)}}$;
and
$x{\dfrac{dx}{dt}=-(y\dfrac{dy}{dt}+z\dfrac{dz}{dt}})$
Work Step by Step
Given: $s={\sqrt{(x^2+y^2+z^2)}}$
Now, on differentiating both sides, we have:
$\dfrac{ds^2}{dt}=\frac{d{{(x^2+y^2+z^2)}}}{dt}$
or, $2s\dfrac{ds}{dt}=2x{\dfrac{dx}{dt}+2y\dfrac{dy}{dt}+2z\dfrac{dz}{dt}}$
or, $\dfrac{ds}{dt} = \frac{x{\frac{dx}{dt}+y\frac{dy}{dt}+z\frac{dz}{dt}}}{\sqrt{(x^2+y^2+z^2)}}$
Now, take $x$ as a constant, then
$s\dfrac{ds}{dt}=y\dfrac{dy}{dt}+z\dfrac{dz}{dt}$
or, $0=x{\frac{dx}{dt}+y\frac{dy}{dt}+z\frac{dz}{dt}}$
$x{\frac{dx}{dt}=-(y\frac{dy}{dt}+z\frac{dz}{dt}})$
Hence,
$\dfrac{ds}{dt} = \dfrac{x{\frac{dx}{dt}+y\frac{dy}{dt}+z\frac{dz}{dt}}}{\sqrt{(x^2+y^2+z^2)}}$
$\dfrac{ds}{dt} = \dfrac{(y\frac{dy}{dt}+z\frac{dz}{dt})}{\sqrt{(x^2+y^2+z^2)}}$;
and
$x{\dfrac{dx}{dt}=-(y\dfrac{dy}{dt}+z\dfrac{dz}{dt}})$
