Find the general integral of…

Find the general integral of $\left(\dfrac1y\sin\dfrac{x}{y}-\dfrac{y}{x^2}\cos\dfrac{y}{x}+1\right)dx+\left(\dfrac1x\cos\dfrac{y}{x}-\dfrac{x}{y^2}\sin\dfrac{x}{y}+\dfrac{1}{y^2}\right)dy=0$.

Step 1: Group the $\sin(x/y)$ terms

Since $d\left[-\cos\dfrac{x}{y}\right]=\sin\dfrac{x}{y}\left(\dfrac1y\,dx-\dfrac{x}{y^2}\,dy\right)=\dfrac1y\sin\dfrac{x}{y}\,dx-\dfrac{x}{y^2}\sin\dfrac{x}{y}\,dy$, these two terms combine into $d\!\left[-\cos\dfrac{x}{y}\right]$.

Step 2: Group the $\cos(y/x)$ terms

Since $d\!\left[\sin\dfrac{y}{x}\right]=\cos\dfrac{y}{x}\left(\dfrac1x\,dy-\dfrac{y}{x^2}\,dx\right)=-\dfrac{y}{x^2}\cos\dfrac{y}{x}\,dx+\dfrac1x\cos\dfrac{y}{x}\,dy$, these two terms combine into $d\!\left[\sin\dfrac{y}{x}\right]$.

Step 3: Group the remaining terms

The leftover terms are $dx+\dfrac{1}{y^2}\,dy=d\!\left[x-\dfrac1y\right]$.

Step 4: Combine all three differentials

The whole left side is

\[d\!\left[-\cos\frac{x}{y}+\sin\frac{y}{x}+x-\frac1y\right]=0.\]