1) How do I find the intersection of f[c_] := y^2 - 2 x^2 (x + 3) == c ((y + 1)^2 (y + 9) - x^2) between the plots of f[0] and f[2]
I tried plugging-in numbers. But I can't seem to find the intersections
2) Also how do I graph this https://i.sstatic.net/OvS9n.jpg ?
I figured out the equation to be y=x^2-9 but I don't know what to do next.
I am uncertain exactly what is wanted. If it is visualization you can use ContourPlot:
fun1[_][x_, y_] := y^2 - 2 x^2 (x + 3); fun2[c_][x_, y_] := c ((y + 1)^2 (y + 9) - x^2); cp = ContourPlot[ Evaluate[(fun1[#][x, y] == fun2[#][x, y]) & /@ {0, 2}], {x, -10, 10}, {y, -10, 10}, MeshFunctions -> (fun2[2][#, #2] &), Mesh -> {{0.}}, MeshStyle -> {PointSize[0.015], Red}, ContourStyle -> {Green, Purple}, PlotLegends -> "Expressions"] 
The mesh function exploits when c=0 the second function is 0.
If an approximation of the points of intersection is desired you can extract them from graphic: e.g.
ind = Cases[cp, Point[x__] :> x, Infinity]; ans = cp[[1, 1, 1]][[First@ind]] yielding:
{{2.63803, -8.8862}, {0.441098, -1.15861}, {0.335547, -0.880514}, \ {-0.376551, -0.864387}, {-0.539491, -1.19441}, {-2.71652, -2.02988}, \ {-2.9964, -0.00107654}, {2.64087, -8.88768}, {-2.7232, -2.03298}, \ {-2.99633, -0.00098642}, {-0.369225, -0.870458}, {0.327706, \ -0.882296}, {0.439565, -1.15777}, {-0.513967, -1.18753}} I re-stress approximates.
If you want to "find":
red = N@Reduce[(fun1[#][x, y] == fun2[#][x, y]) & /@ {0, 2}, {x, y}] yields:
(x == -3. || x == -2.72156 || x == -0.537205 || x == -0.378744 || x == 0.340628 || x == 0.441236 || x == 2.64504 || x == -2.8947 - 2.63671 I || x == -2.8947 + 2.63671 I) && y == 2.99448*10^-8 (-1.65309*10^7 + 162. x - 1.00263*10^8 x^2 - 3.40627*10^7 x^3 + 6.49485*10^6 x^4 + 4.1719*10^6 x^5 + 638632. x^6 - 19352. x^7 - 3200. x^8) A way to extract this result (real solutions):
xval = Cases[red, x == r_ :> r, Infinity]; yval[u_] := First[Cases[red, y == r_ :> r]] /. x -> u; calc = {#, yval@#} & /@ Cases[xval, _Real] yielding:
{{-3., 0.}, {-2.72156, -2.03094}, {-0.537205, -1.19225}, {-0.378744, \ -0.867192}, {0.340628, -0.88046}, {0.441236, -1.15756}, {2.64504, \ -8.88754}} Comparing with the approximations the original plot and the above real solutions are overlayed with an "x":
cp2 = Show[cp, ListPlot[calc, PlotMarkers -> {"\[Times]", 20}]] 
If this is not what is desired, my apologies.
For finding the intersections, never discount the simplest way to do it,
pts = {x, y} /. NSolve[{f[0], f[2]}, {x, y}, Reals] (* {{2.64504, -8.88754}, {-3., 0.}, {0.441236, -1.15756}, {0.340628, -0.88046}, {-0.537205, -1.19225}, {-0.378744, -0.867192}, {-2.72156, -2.03094}} *) And you can see that you got the points correctly by plotting,
Show[ContourPlot[Evaluate[{f[0], f[2]}], {x, -5, 5}, {y, -10, 10}], ListPlot[pts, PlotStyle -> Directive[PointSize[Large], Red]] ] 
And that other plot, you can do something like this
func = #^2 - 9 &; ParametricPlot[{ {x, func[x]}, {x, -func[x]}, {func[x], x}, {-func[x], x}, {Sqrt[-x^2 + 9], x}, {-Sqrt[-x^2 + 9], x} } , {x, -10, 10}, PlotRange -> {{-10, 10}, {-10, 10}}] 
x^2andSqrt[x]might help. In the end you might have that $\endgroup$