The equation can be divided through by 2: x^2+6x+8y+1=0; x^2+6x+9-9+1+8y=0; (x+3)^2-8+8y=0; -8(y-1)=(x+3)^2. The vertex is at (-3,1) and the curve is a parabola like an inverted U. The vertex line is x=-3, and the focus lies on this line. The properties of a parabola make the following true: (y-d)^2=(x-h)^2+(y-k)^2, where (h,k) is the focus and y=d is the directrix. The rule is that all points on the parabola are equidistant from the focus and directrix line (shortest distance is a perpendicular). So, (y-d)^2-(y-k)^2=(x-h)^2; (y-d-y+k)(y-d+y-k)=(x-h)^2; (k-d)(2y-k-d)=(x-h)^2; 2y(k-d)-(k-d)(k+d)=(x-h)^2. comparing terms we see that (x+3)^2=(x-h)^2, so h=-3; 2y(k-d)=-8y, so k-d=-4; -(k-d)*(k+d)=8, so 4(k+d)=8; k+d=2. Add together the equations containing k and d: 2k=-2, k=-1 and d=3. So the focus is (h,k)=(-3,-1) and the directrix is y=3.
Check: (y-3)^2=(x+3)^2+(y+1)^2; (y-3)^2-(y+1)^2=(x+3)^2; -4(2y-2)=-8(y-1)=(x+3)^2 as above. Seems all OK.
To draw the graph mark the vertex of the upside down U. That's the highest point. Also draw the vertex line x=-3, which is the axis of symmetry. When y=0 (x axis) x=-3+2sqrt(2)=0.17 approx and x=-3-2sqrt(2)=-5.83, so that's where the curve intersects the x axis. When x=0 y=-0.125.
You can also draw the directrix line at y=3 which lies above the curve and the focus which is on the inside of the U.