Question 694400
You probably meant 
{{{9x^2+25y^2=1}}}
and want to express it in the form
{{{x^2/a^2+y^2/b^2=1}}},
which is appropriate for an ellipse centered at the origin, with axes along the x- and y-axes.
 
You could write your equation as
{{{x^2/(1/9)+y^2/(1/25)=1}}} or as {{{x^2/(1/3)^2+y^2/(1/5)^2=1}}}
 
Even without re-writing the equation, it was obvious that it was the equation for an ellipse centered at the origin, with axes along the x- and y-axes, because there were no terms in {{{x}}} or {{{y}}}, or {{{xy}}}.
 
VERTICES:
The vertices, then, are the intersections with the x- and y-axes, were {{{y=0}}} and {{{x=0}}}, and those are points you need to graph the ellipse.
{{{y=0}}} --> {{{9x^2=1}}} --> {{{x^2=1/9}}} --> {{{x =1/3}}} or {{{x=-1/3}}} gives you vertices (-1/3,0) and (1/3,0).
{{{x=0}}} --> {{{25y^2=1}}} --> {{{y^2=1/25}}} --> {{{y=1/5}}} or {{{x=-1/5}}} gives you co-vertices (0,-1/5) and (0,1/5).
 
AXES:
Vertices (-1/3,0) and (1/3,0) are farther from center (0,0) than co-vertices (0,-1/5) and (0,1/5):
{{{1/3>1/5}}},
so the segment between (-1/3,0) and (1/3,0) is called the major axis,
and the segment/distance from each of those vertices to the center is called the semi-major axis, represented as {{{a}}},
so {{{a=1/3}}}.
The distance from the center to the co-vertices is the semi-minor axis:
{{{b=1/5}}}.
 
FOCI:
The foci would be useful to draw the ellipse if you wanted a very accurate representation.
They are points on both semi-major axes at a distance from the center called the focal distance, represented as {{{c}}}.
Since for all points of the ellipse the sum of the distances to the foci is constant, the same.
For the vertices, the distance to the nearest focus is {{{a-c}}} and the distance to the other one is {{{a+c}}}, so the sum is {{{2a}}}.
For the co-vertices, the distance to each focus in the hypotenuse of a right triangle with leg lengths {{{b}}} and {{{c}}}.
Each of those distances is {{{sqrt(b^2+c^2)}}}, so the sum is {{{2sqrt(b^2+c^2)}}}.
Since that should be the same as for the vertices,
{{{2a=2sqrt(b^2+c^2)}}} <--> {{{a=sqrt(b^2+c^2)}}} --> {{{highlight(a^2=b^2+c^2)}}}
That formula allows you to find the focal distance and graph the foci.
{{{(1/3)^2=(1/5)^2+c^2}}} --> {{{1/9=1/25+c^2}}} --> {{{c^2=1/9-1/25}}} --> {{{c^2=(25-9)/225}}} --> {{{c^2=16/225}}} --> {{{highlight(c=4/15)}}}
So the foci will be at (-4/15,0) and (4/15,0).
 
GRAPHING:
We can plot the vertices co-vertices and foci, like this
{{{drawing(300,300,-0.4,0.4,-0.4,0.4,
arrow(0,0,0,0.4),arrow(0,0,0,-0.4),
arrow(0,0,0.4,0),arrow(0,0,-0.4,0),
blue(circle(-1/3,0,0.01)),blue(circle(1/3,0,0.01)),
red(circle(-4/15,0,0.01)),red(circle(4/15,0,0.01)),
green(circle(0,-0.2,0.01)),green(circle(0,0.2,0.01)),
triangle(-4/15,0,4/15,0,0,0.2),rectangle(0,0,0.03,0.03),
locate(0.37,-0.03,x),locate(0.03,0.4,y),
locate(0,0.2,co-vertex),locate(0,-0.2,co-vertex),
locate(-0.4,-0.16,vertex),locate(0.28,-0.16,vertex),
locate(-0.27,0.2,focus),locate(0.26,0.2,focus),
arrow(-1/3,-0.15,-1/3,-0.02),arrow(1/3,-0.15,1/3,-0.02),
arrow(-4/15,0.15,-4/15,0.02),arrow(4/15,0.15,4/15,0.02)
)}}} Then draw the ellipse {{{graph(300,300,-0.4,0.4,-0.4,0.4,9x^2+25y^2<1)}}}
I would just draw a curve that passes through vertices and co-vertices and looks  like it could be an ellipse.
To have a real ellipse, you would have to stick pins at the foci; make a loop of thread with a total length (all around) equal to the the distance between the foci, plus the distance between the vertices; throw the loop over the pins; stretch the loop with the tip of a pencil, and draw around keeping the loop fully stretched.