Any irreducible cubic affine curve has at most 1 singular point Let $k$ be a fields (it may be assumed that $k$ is algebraically closed if necessary). Let $f\in k[x,y]$ be an irreducible cubic affine curve. I need to prove that $f$ had at most one singular point. I need an elementary proof not dependent on Bezout`s theorem. I made a trial: 
Notice first that the assumption that $f$ is irreducible guarantees that $f$ has only finitely many singular points. If $P=(x_0,y_0)$ is a singular point of $f$ then, since $f(a,b)=0$, $f$ must be of the form: 
\begin{align}
f= (x-x_0)u+ (y-y_0)v,
\end{align}
where $u,v\in k[x,y]$. Thanks to $\frac{\partial f}{\partial x}(P)=0= \frac{\partial f}{\partial y}(P)$, $P$ is a singular point of $f$ is equivalent to saying that $P\in u\cap v$. Thus, 
\begin{align}
u&=(x-x_0)\alpha + (y-y_0)\beta, \\
v&= (x-x_0) \widetilde{\alpha} + (y-y_0) \widetilde{\beta}
\end{align}
for some $\alpha,\beta, \widetilde{\alpha},\widetilde{\beta}\in k[x,y]$. So, we must have:
\begin{align}
f=(x-x_0)^2 A+(x-x_0)(y-y_0)B+ (y-y_0)^2C 
\end{align}
for $A,B,C\in k[x,y]$. Because $\text{deg }f=3$, then at least one of $A,B,$ and $C$ is of degree 1. Without loss of generality, assume that $\text{deg }f=1$. Hence, $A(x,y)=m_1x+m_2y+m_3$ with $0\not= m_1\in k$. Let $Q=(s,t)\in k^2$ be another singular point. Then, by definition, $\mu_Q(f)>1$, where $\mu_Q(f)$ is the multiplicity of $f$ at $Q$. But, also by definition, $\mu_Q(f)=\mu_0(g)$, where $0=(0,0)$ and $g$ is the curve obtained from $f$ after shifting the coordinates $(x,y)$ to $(x',y')=(x-s,y-t)$. That is,
\begin{align}
g(x', y')=f(x'+s,y'+t)&=(x'+s-x_0)^2 \widetilde{A} +(x'+s-x_0)(y'+t-y_0) \widetilde{B} 
\\&+ (y'+t-y_0)^2 \widetilde{C},
\end{align}
where $\widetilde{A},\widetilde{B},$ and $\widetilde{C}$ are the curves after shifting the coordinates. But, definitely, $g$ contains a term of degree less than 2 which means $\mu_Q(f)=\mu_0(g)<2$, contradicting $\mu_Q(f)>1$ as claimed. 
I really appreciate if someone can read and revise this solution. 
 A: Your proof looks roughly correct, but it's a pain to read because you haven't fully embraced the reality of your situation (indeed, I didn't read it carefully all the way through because of how notationally unappetizing it was). If you're going to compute in coordinates, pick good coordinates! Your goal should be to make the calculations as simple as possible, both for yourself as well as your readers. Here's a proof of the statement which does that:
Proof: Suppose $f$ is an irreducible cubic so that $V(f)$ has at least two singular points. Up to a linear change of coordinates, we may assume that two of these singular points are located at $(0,0)$ and $(0,1)$. (We assume $k$ algebraically closed here.)
From the assumption that $(0,0)$ is on the curve, we get that $f$ has no constant term. From the assumption that $\frac{df}{dx}=\frac{df}{dy}=0$ at $(0,0)$, we get that the linear terms of $f$ vanish as well. At this point we can write $$f=ax^3+bx^2y+cxy^2+dy^3+px^2+qxy+ry^2$$
From the assumption that $(0,1)$ is on the curve, we get that $d+r=0$. From the assumption that $\frac{df}{dy}=0$ at $(0,1)$ we get that $3d+2r=0$. So $d=r=0$. Now $f$ is divisible by $x$, which contradicts the assumption that $f$ was to be irreducible. $\blacksquare$
Commentary: This is basically the same strategy as the proof using Bezout's theorem. That proof says "look at the line through the singular points: it meets the curve at 4 points, so by Bezout it must be a component of our curve, and so $f$ is reducible." This proof says "look at the line through the singular points: we play around with some coefficents, and then the line through those points must be a component of our curve, and so $f$ is reducible." Even when you decide not to use a result, you may still be able to use a very similar proof as you would have if you had access to that result.
