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If 5x+5y= 5x+y then prove that dy/dx + 5y-x​= 0

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Are you sure that the question is correct??because from this equation,we get y=0
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yes it is correct
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Differentiating , we get 
5xlog 5   + 5y log 5 dy/dx =  5(x+y) log 5 ( 1 + dy/dx)

ie.,     (5y - 5(x+y) )dy/dx = 5(x+y) - 5x
or dy/dx = 5(x+y) - 5x / ​(5y - 5(x+y) )   ..........(1)
But 5x + 5y = 5(x+ y)     (given)
Applying this on (1) , we get
dy/dx = 5x + 5y  - 5x / 5y - ( 5x + 5y)       =      - 5y/5x
re arranging , dy/dx + 5y/5x = 0 
or dy/dx + 5(y-x)  = 0       (as 5y/5x = 5(y-x) )
Hope that helps.
 

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can you please explain the first step?how did you get 5xlog5 + 5ylogy dy/dx??
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Differentiation of ax wrt x= axloga 
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how did you get (1+ dy/dx) in first step
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Here , we have differentiated 5(x+y) . 
Let z = 5(x + y)
Let's differentiate using chain rule 
dz/dx   =   d/dx ( 5(x+y) ) = 5(x+y)log 5 d/dx ( x+ y)     ( d/dx ( ax )  = ax loga )
=  5(x+y)log 5 ( d/dx ( x)  + d/dx(y) )  =  5(x+y)log 5 ( 1 + dy/dx) .                     ( d/dx (x)= 1  and d/dx(y) = dy/dx )
Hope that helps and all the best !
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thank you........@ RS Prasobh sankar
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