Answer
$\triangle T \approx -0.01648$ mg and tension decreases
Work Step by Step
The differential form can be evaluated as follows:
$dT=\dfrac{\partial T}{\partial r} dr + \dfrac{\partial T}{\partial R} dR$
Take the partial derivatives.
$dT=[\dfrac{-4mgRr}{(2r^2+R^2)^2} ]dr + [\dfrac{mg(2R^2-r^2)}{(2r^2+R^2)^2} ] dR$
Re-arrange as: $\triangle T \approx [\dfrac{-4mgRr}{(2r^2+R^2)^2} ] \triangle r + [\dfrac{mg(2R^2-r^2)}{(2r^2+R^2)^2} ] \triangle R$
Plug in the given data and we have
$\triangle T \approx [\dfrac{-4mg(3) \times (0.7)}{(2(0.7)^2+(3)^2)^2} ] \times (0.1) + [\dfrac{mg(2(0.7)^2-(3)^2)}{(2(0.7)^2+(3)^2)^2} ] \times (0.1)$
Hence, we have $\triangle T \approx -0.01648$ mg
The tension is negative and decreases.