Obtaining equation of Tangent(s) to A Curve at origin by equating the lowest degree terms to zero. In a rational algebraic expression in 2 variable having no constant term, equation of the tangent at origin can be formed by equating the lowest degree term to be zero.
Ex. For $ x^4+y^4+2xy^2-2y=0 $ the equation of tangent at origin is $ 2y=0 $ or $ y=0 $
Is there an Analytical Method to prove this or is this pure observation?
 A: I assume that you are writing about implicit equations that are polynomial in the two variables $x$ and $y$. Let's also assume that the curve does go through the origin and that there is at least one tangent line to the curve through the origin.
With those assumptions, I can give you an argument that removing all but the lowest-degree terms will result in another polynomial implicit equation with the same tangent line(s). This is not a proof, but it may be convincing. I'll argue from your given example, which may be clear and convincing enough. So we start with
$$x^3+y^3=3xy$$
To look at the tangent lines to the curve through the origin we are concerned about very small values of $x$ and of $y$--larger values do not affect the tangent lines. The lowest degree term(s) in our sample equation, namely $3xy$, has the degree two. Therefore let divide both sides of the equation by a polynomial in one variable with the same degree, namely $x^2$. Doing that we get
$$x + y\left(\frac yx\right)^2 = 3\left(\frac yx\right)$$
As we let $x$ and $y$ approach zero on the curve close to a non-vertical tangent line, then the slope of that tangent line is finite and well-defined and $\frac yx$ approaches that slope. Thus all the terms with $\frac yx$ also have well-defined limits. So taking the limits of both sides of that equation as $x$ and $y$ approach zero, we get
$$0 + 0\left(\frac yx\right)^2 = 3\left(\frac yx\right)$$
or equivalently
$$0 = 3\left(\frac yx\right)$$
Now we multiply the the previously-divided-by expression $x^2$ and we get
$$0=3xy$$
You see that, in effect, we removed the higher-degree terms from the equation and got an equation that is true at the limit as $x$ and $y$ approach zero. So the tangent lines to the last equation are the same as those for the original equation.
There is a major weakness in that argument--what if the tangent line is vertical, so $\frac yx$ does not have a finite limit as $x$ and $y$ approach zero? We can handle that by dividing both sides of the original equation by $y^2$ (rather than the $x^2$ that we used) and note that the expression $\frac xy$ approaches zero as $x$ and $y$ approach zero on the curve close to the vertical tangent line. Doing that we get the equation at the limit
$$0 = 3\left(\frac xy\right)$$
and still end up with
$$0=3xy$$
Note that there were multiple assumptions at the start and that our reasoning was not rigorous. Also be aware that not all implicit function that have a point at the origin have any tangent lines. An example is
$$x^2+y^2=x^2y^2$$
though you could argue that dropping all but the lowest-degree terms does yield an equation that shows the lack of tangent lines.
