This is a follow-up on the following answer: https://math.stackexchange.com/a/4687845/755010
To the question of how to graph $y^{4}x^{3}+x^{4}-y=0$ without a computer, user Somos suggested to approximate it by easy to graph equations $y^4+x=0$, $(xy)^3-1=0$, and $x^4-y=0$. This is nicely pictured in user Jean Marie's screenshot: 
It seems to me that this technique could be more generally applied to a variety of difficult to solve quartics. Would there be a general theorem on the conditions under which we could achieve satisfactory results with this technique?
If we look at $y^4x^3+x^4z^3-yz^6=0$ the projective curve, it has singular points at $(1:0:0)$ and $(0:1:0)$ which both are at the line at infinity $z=0$.
So let's investigate the behavior around the singular points:
$(y/x)^4+(z/x)^3-(y/x)(z/x)^6=0$ has $(z/x)^3=0$ as tangent cone and can be approximated by the terms of lowest degree $(y/x)^4+(z/x)^3=0$ or $(y/x)^4=-(1/x)^3=0$ or $y^4=-x.$
$(x/y)^3+(x/y)^4(z/y)^3-(z/y)^6=0$ has $(x/y)^3=0$ as tangent cone and can be approximated by the terms of lowest degree $(x/y)^3=(1/y)^6$ or $x^3y^3=1.$
$(y/z)^4(x/z)^3+(x/z)^4-y/z=0$ is the original curve and has tangent cone $(y/z)=0$ at $(0:0:1)$ which is on the curve, and can be approximated by the terms of lowest degree $x^4=y.$
Added: More generally, if you have a point $(a:b:1)$ on the curve, in your affine, translate the point to the origin by $(y+b)^4(x+a)^3+(x+a)^4-(y+b)=0$ and collect lowest terms to get the tangent line $y-b=\frac{4a^3+3a^2b^4}{1-4a^3b^3} (x-a),$ or $$3ab^4(x-a)^2+12a^2b^3(x-a)(y-b)+(4a^3+3a^2b^4)(x-a)-(1-4a^3b^3)(y-b)=0$$ if you keep the degree less than three terms.
Now, the curve might not have more rational points than the three we investigated. Also note that the conic we found is not the osculating conic, and has lower contact than the five we expect for those. Next, we could search for sextactic points.