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asking this question few days ago, I first saw that "trick" in the given answer. that trick enabled me to solve the question pretty easily (not the solution given), by adding the 2 equations: $$(1):\frac{dx}{dz} = \frac{y+xz}{z^2-1}$$ $$(2):\frac{dy}{dz} = \frac{x+yz}{z^2-1}$$ $$(1)+(2): \frac{d(x+y)}{dz} = \frac{(x+y)(z+1)}{z^2-1}$$ and from there it was easy to solve the question.

However I still wonder if that was "legal" move to assume that:

$$\frac{dx}{dz} + \frac{dy}{dz} = \frac{d(x+y)}{dz}$$

I have to wonder what is the meaning of this ? in what circumstances that legal move ?

Community
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d_e
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    you are leveraging the linearity of the differential operator so $\dot{(x+y)} = \dot{x} + \dot{y}$ – Chinny84 Oct 27 '15 at 15:00

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This is always a legal move. Part of what the "linearity of the derivative" tells us is that $$ \frac{d}{dz}(x(z) + y(z)) = \frac{dx}{dz} + \frac{dy}{dz} $$ for any differentiable functions $x(z),y(z)$.

Ben Grossmann
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