Transform to polar coordinates x = r cos phi, y = r sin
Transform to polar coordinates $x = r\cos\varphi,\ y = r\sin\varphi$: 1° $u = xy'-y$; 2° $y' = \dfrac{x+y}{x-y}$.
Explanation of Method
Rewriting a derivative expression and a differential equation in terms of r and φ.
Step 1: Preliminary: $y'_{x}$ in polar form
Treat $r$ as a function of $\varphi$, so that the curve is given parametrically by $\varphi$. Then
\[ \frac{dx}{d\varphi} = r'\cos\varphi - r\sin\varphi, \qquad \frac{dy}{d\varphi} = r'\sin\varphi + r\cos\varphi, \]
where $r' = \dfrac{dr}{d\varphi}$. Dividing one by the other gives the master formula
\[ y'_{x} = \frac{dy/d\varphi}{dx/d\varphi} = \frac{r'\sin\varphi + r\cos\varphi}{r'\cos\varphi - r\sin\varphi}. \tag{$*$} \]
Step 2: Part 1°: substitute into $u = xy'-y$
\[ u = r\cos\varphi\cdot\frac{r'\sin\varphi+r\cos\varphi}{r'\cos\varphi-r\sin\varphi} - r\sin\varphi = \frac{r\cos\varphi\bigl(r'\sin\varphi+r\cos\varphi\bigr)-r\sin\varphi\bigl(r'\cos\varphi-r\sin\varphi\bigr)}{r'\cos\varphi-r\sin\varphi}. \]
Step 3: The numerator collapses
Expanding, the two $rr'\sin\varphi\cos\varphi$ terms cancel and the two remaining terms combine by the Pythagorean identity: \[rr'\sin\varphi\cos\varphi + r^{2}\cos^{2}\varphi - rr'\sin\varphi\cos\varphi + r^{2}\sin^{2}\varphi = r^{2}\bigl(\cos^{2}\varphi+\sin^{2}\varphi\bigr) = r^{2}.\]
\[u = \dfrac{r^{2}}{r'\cos\varphi - r\sin\varphi}\]
Step 4: Part 2°: rewrite the right-hand side
Both $x+y$ and $x-y$ carry a common factor $r$, which cancels: \[\frac{x+y}{x-y} = \frac{r\cos\varphi+r\sin\varphi}{r\cos\varphi-r\sin\varphi} = \frac{\cos\varphi+\sin\varphi}{\cos\varphi-\sin\varphi}.\]
Step 5: Equate with $(*)$ and cross-multiply
\[ \frac{r'\sin\varphi+r\cos\varphi}{r'\cos\varphi-r\sin\varphi} = \frac{\cos\varphi+\sin\varphi}{\cos\varphi-\sin\varphi} \]
\[ \bigl(r'\sin\varphi+r\cos\varphi\bigr)\bigl(\cos\varphi-\sin\varphi\bigr) = \bigl(\cos\varphi+\sin\varphi\bigr)\bigl(r'\cos\varphi-r\sin\varphi\bigr). \]
Step 6: Expand and collect
Moving everything to the left and grouping the $r'$ terms separately from the $r$ terms: \[r'\bigl(\sin\varphi\cos\varphi-\sin^{2}\varphi-\cos^{2}\varphi-\sin\varphi\cos\varphi\bigr) + r\bigl(\cos^{2}\varphi-\sin\varphi\cos\varphi+\sin\varphi\cos\varphi+\sin^{2}\varphi\bigr) = 0.\]
The first bracket is $-1$ and the second is $+1$, so the whole equation reduces to $-r'+r=0$.
Step 7: Verification
\[\frac{dr}{d\varphi} = r \;\Longrightarrow\; r = Ce^{\varphi} \quad\text{(the equiangular, or logarithmic, spiral)}\]
\[\text{For } C=1:\ r=e^{\varphi} \text{ is exactly the curve } \arctan\frac{y}{x}=\ln\sqrt{x^{2}+y^{2}} \text{ of Problems 217 and 225} \;\;\checkmark\]
\[\text{Symbolic check of 2}^{\circ}\text{: the residual } y'_{x}-\frac{x+y}{x-y} \text{ has numerator proportional to } \bigl(r-r'\bigr) \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,u = \dfrac{r^{2}}{r'\cos\varphi - r\sin\varphi}\,}\]
Final answer:
\[\boxed{\,\textbf{1}^{\circ}\ \ u = \dfrac{r^{2}}{r'\cos\varphi-r\sin\varphi}, \qquad \textbf{2}^{\circ}\ \ \dfrac{dr}{d\varphi} = r\,}\]
Remark
Remark. Part 1° has a clean geometric meaning: $xy'-y$ is (up to sign) the $y$-intercept of the tangent line, and the polar form says this quantity equals $r^{2}$ divided by $dx/d\varphi$. Part 2° shows how dramatically polar coordinates can simplify a differential equation: an awkward homogeneous equation in $x$ and $y$ becomes the simplest growth equation there is, $r' = r$.