$\newcommand\C{\mathcal C}$Let $\C:f(x,y)=0$ be a real plane algebraic curve, we will also assume that $\C$ has real points and in particular that $P=(0,0)$ is a point of $\C$. Let $f_d(x,y)$ be the homogeneous part of the lowest degree of $f(x,y)$ such that $P$ is an isolated (real) point of $\C_d : f_d(x,y)=0$. Can I conclude that $P$ is an isolated point of $\C$?
If the answer above is yes, then can I make a similar statement for real analytic curves: so $f(x,y)$ is not anymore a bivariate polynomial but an analytic function and $f_d(x,y)$ is the lowest order non-trivial term in the Taylor series expansion of $f$ at $P$ (it is a homogeneous polynomial, so $\C_d$ remains algebraic).
The answer to the first question is yes. A point $p$ of a real plane curve $C$ is isolated in $C(\Bbb R)$ iff all the tangent directions at $p$ are non-real (the tangent directions are the fiber of the strict transform of $C$ over $p$ in the blowup $Bl_p\Bbb A^2$). These tangent directions are determined by the lowest term of the Taylor expansion of $f$ at $p$, so going from $f_d$ to $f$ doesn't change the tangent directions, and thus $p$ is isolated as a point of $V(f_d)(\Bbb R)$ iff it's isolated as a point of $V(f)(\Bbb R)$.
The extension to the real-analytic case should proceed in the same fashion, except you'll need to extend $f$ to a complex analytic function in some open neighborhood of $\Bbb R^2\subset \Bbb C^2$ and keep track of the real points.