Suppose that $f(x,y)$ is a polynomial with coefficients in the complex numbers such that $f(x,y)=0$ defines a non-linear variety of dimension 1.
How could I show that there is no $\alpha\in\mathbb{C}$ such that $\dfrac{\partial f}{\partial x}+\alpha\dfrac{\partial f}{\partial y}$ is not identically equal to zero?
My attempt:
Suppose otherwise. If $\alpha=0$ then from the equation we would have $\dfrac{\partial f}{\partial x}=0$, and so $f(x,y)=g(y)$ for some polynomial $g(y)\in \mathbb{C}[y]$, but in this case the equation $f(x,y)=g(y)=0$ defines a finite union of varieties given by equations of the form $y=z_i$ for some $z_1,\ldots,z_m\in\mathbb{C}$.
Now, we can write the polynomial in the form $\displaystyle{f=\sum_{k=0}^n g_k(y)\cdot x^k}$ where $g_k(y)\in \mathbb{C}[y]$. Making the calculations we have that
\begin{align*} \dfrac{\partial f}{\partial x}+\alpha\dfrac{\partial f}{\partial y}=\alpha\cdot g_n'(y)\cdot x^n+\left(\sum_{k=0}^{n-1} \left((k+1)\cdot g_{k+1}(y)+\alpha\cdot g'_k(y) \right)\cdot x^k\right) \end{align*} and if it is equal to the zero polynomial then we would have:
$$\begin{cases}g'_n(y)=0\\ (k+1)\cdot g_{k+1}(y)+\alpha\cdot g'_k(y)=0\end{cases}$$
This implies that the polynomials $g_k(y)$ have degree $n-k$ for $k=0,1,\ldots,n$ (as one can find them recursively), but this does not take me to a contradiction.
On the other hand, if we make the calculations with the polynomial $f(x,y)=ax^2+bxy+cy^2$, it turns out that $\dfrac{\partial f}{\partial x}+\alpha\dfrac{\partial f}{\partial y}\equiv 0$ if and only if $b^2=4ac$, in which case we would have that $f(x,y)=\left(ax+\dfrac{b}{2a}\right)^2$.
The latter make me think that perhaps one has to use irreducibility of the non-linear variety, but I still do not know how to argue.