Answer
$\dfrac{1}{(1+uvw)^2}(-v^2w du+dv-uv^2dw)$
Work Step by Step
Given the function $T=\dfrac{v}{1+uvw}$
The differential form can be evaluated as follows:
$dT=\dfrac{\partial T}{\partial u} du +\dfrac{\partial T}{\partial v} dv+ \dfrac{\partial T}{\partial w} dw$
We need to find the partial derivatives w.r.t. $u$ ; $v$ and $w$ as follows:
$T_u=\dfrac{-v(vw)}{(1+uvw)^2} \\T_v=\dfrac{1+uvw-v(uw)}{(1+uvw)^2} \\T_w=\dfrac{-v(uv)}{(1+uvw)^2}$
Hence, we have
$dT=\dfrac{-v(vw)}{(1+uvw)^2}du+\dfrac{1+uvw-v(uw)}{(1+uvw)^2} dv+\dfrac{-v(uv)}{(1+uvw)^2} dw=\dfrac{1}{(1+uvw)^2}(-v^2w du+dv-uv^2dw)$
