What is the solution for derivative of 1/x+1/y =1 ?

The solution for derivative of 1/x+1/y =1 is y=-x/(1-x^2-c_{1)x}

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derivative of 1/x+1/y =1

\frac{d}{d x} (\frac{1}{x} + \frac{1}{y}) = 1

y = - \frac{x}{1 - x ^{2} - c _{1} x}

Solution steps

\frac{d}{d x} (\frac{1}{x} + \frac{1}{y}) = 1

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

\frac{1}{x} + \frac{1}{y} = \int 1 d x

\int 1 d x = x + c_{1}

\frac{1}{x} + \frac{1}{y} = x + c_{1}

Isolate

y : y = - \frac{x}{1 - x ^{2} - c _{1} x}

y = - \frac{x}{1 - x ^{2} - c _{1} x}

\frac{d}{d x} (e^{x} sin (y)) = 0

\frac{d y}{d t} = \frac{t}{t ^{2} y - y}

y^{^{'}} = (cos (x)) sin (y)

\frac{d}{d θ} (y (θ)) y (θ) = cos (e^{- 5 θ^{5}})

4 t \frac{d y}{d t} + 22 y = 0

What is the solution for derivative of 1/x+1/y =1 ?
The solution for derivative of 1/x+1/y =1 is y=-x/(1-x^2-c_{1)x}