Properties of Line Integral

[WMDhamnekar]

TL;DR Summary

Let f (x, y) and g(x, y) be continuously differentiable real-valued functions in a region R.
Show that $\displaystyle\oint_C f\nabla g \cdot dr = -\displaystyle\oint_C g\nabla f \cdot dr$ for any closed curve in R

I don't have any idea to answer this question. So, any math help will be accepted.
I know $\nabla fg = f\nabla g + g\nabla f \rightarrow (1)$ But I don't understand to how to use (1) here?

 

[malawi_glenn]

What will $\displaystyle\oint_C \vec \nabla (fg) \cdot d \vec r$ become?

 

[WMDhamnekar]

What will $\displaystyle\oint_C \vec \nabla (fg) \cdot d \vec r$ become?

I think $\displaystyle\oint_C \nabla (fg) \cdot dr = \displaystyle\oint_C f\nabla g \cdot dr + \displaystyle\oint_C g\nabla f \cdot dr$ for any closed curve.

But, for any closed curve in R , $\displaystyle\oint_C \nabla (fg) \cdot dr = 0,$


Hence, we can deduce that $\displaystyle\oint_C f\nabla g \cdot dr = -\displaystyle\oint_C g\nabla f \cdot dr$
Is that logic correct?

 

[malawi_glenn]

I think ∮C∇(fg)⋅dr=∮Cf∇g⋅dr+∮Cg∇f⋅dr for any closed curve.

This is valid for all lines.