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$\frac{\delta(x-y)^2}{\delta x}=\frac{\delta(x+y-1)}{\delta x}$

I did:

$$\frac{\delta(x-y)^2}{\delta x}=\frac{\delta(x+y-1)}{\delta x} \Leftrightarrow \\ 2(x-y)(\frac{\delta (x-y)}{\delta x}) = 1 + \frac{\delta y}{\delta x} \Leftrightarrow \\ 2(x-y)(1-\frac{\delta y}{\delta x}) = 1+ \frac{\delta y}{\delta x} \Leftrightarrow \\ (2x-2y)(1-\frac{\delta y}{\delta x}) = 1+ \frac{\delta y}{\delta x} \Leftrightarrow \\ (2x-2y)+(2y-2x)\frac{\delta y}{\delta x} = 1+\frac{\delta y}{\delta x} \Leftrightarrow \\ (2y-2x)\frac{\delta y}{\delta x} = 1+\frac{\delta y}{\delta x}-(2x-2y) \Leftrightarrow \\ (2y-2x)\frac{\delta y}{\delta x} = 1+\frac{\delta y}{\delta x}+(2y-2x) \Leftrightarrow \\ ???$$

What do I do next? Did I do it correctly so far?

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    Keep going. Solve for the derivative by moving things around. – 2017-01-16
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    This is confusing. Is $n$ a function of $x$? Is it a constant? How do you get $\frac {d(n+y-1)}{dx}=1+\frac {dy}{dx}$? – 2017-01-16
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    if you replaced dy/dx by A at line 3 of your working - how would you go about the algebra to make A the subject of the expression - i, A = ..... You have to do that same process for dy/dx. Where you are at the end, you need to bring the dy/dx onto the LHS, then divide by a suitable expression on both sides to make dy/dx the subject – 2017-01-16
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    @lulu the $n$ is supposed to be $x$, I'll fix it – 2017-01-16
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    No problem. I edited the text to replace $n$ by $x$ in the header and in the first line. – 2017-01-16
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    Are you clear in the solution now? we have $2(x-y)(1-y')=1+y'$ so you just need to solve for $y'$. – 2017-01-16
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    @lulu Yep, I solved it. Replacing $\frac{\delta y}{\delta x}$ by A did the trick for me. I'll post an answer. – 2017-01-16

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I solved it:

$$\frac{\delta(x-y)^2}{\delta x}=\frac{\delta(x+y-1)}{\delta x} \Leftrightarrow \\ 2(x-y)(\frac{\delta (x-y)}{\delta x}) = 1 + \frac{\delta y}{\delta x} \Leftrightarrow \\ 2(x-y)(1-\frac{\delta y}{\delta x}) = 1+ \frac{\delta y}{\delta x} \Leftrightarrow \\ (2x-2y)(1-\frac{\delta y}{\delta x}) = 1+ \frac{\delta y}{\delta x} \Leftrightarrow \\ (2x-2y)+(2y-2x)\frac{\delta y}{\delta x} = 1+\frac{\delta y}{\delta x} \Leftrightarrow \\ (2y-2x)\frac{\delta y}{\delta x} = 1+\frac{\delta y}{\delta x}-(2x-2y) \Leftrightarrow \\ (2y-2x)\frac{\delta y}{\delta x} = 1+\frac{\delta y}{\delta x}+(2y-2x) \Leftrightarrow \\ $$

Then I replaced $\frac{\delta y}{\delta x}$ by $A$:

$$(2y-2x)A = 1+A+(2y-2x) \Leftrightarrow \\ (2y-2x)A-A = 1+(2y-2x) \Leftrightarrow \\ A(2y-2x-1) = 1 + (2y-2x) \Leftrightarrow \\ A = \frac{2y-2x+1}{2y-2x-1} \Leftrightarrow \\ \frac{\delta y}{\delta x} = \frac{2y-2x+1}{2y-2x-1}$$

Done.