We can write: ${{y}^{2}}(a{{t}^{2}}-bt+c)=d\Rightarrow {{y}^{2}}=\frac{d}{a{{t}^{2}}-bt+c}$ ${{x}^{2}}=\frac{d{{t}^{2}}}{a{{t}^{2}}-bt+c}$ $f(x,y)=\frac{ad{{t}^{2}}+bd}{a{{t}^{2}}-bt+c}=k\Rightarrow (ad-ka){{t}^{2}}+kbt+bd-kc=0$ Just put that the delta it is non negative, and you get the variation of the $k$ so the maximum and the minimum of the fraction.

the smallest value of this function cannot be found

Let us define \(z = \frac{x}{y}\). We have that: \[ax^2 - bxy + cy^2 = d \Rightarrow az^2 - bz + c = \frac{d}{y^2} \iff y^2 = \frac{d}{{az^2 - bz + c}} \text{ and}\]\[ax^2 - bxy + cy^2 = d \Rightarrow a - \frac{b}{z} + \frac{c}{z^2} = \frac{d}{x^2} \iff x^2 = \frac{dz^2}{{az^2 - bz + c}} \Rightarrow\]\[f(x,y) = \frac{d(az^2+c)}{az^2-bz+c} = d \cdot A\]Let the minimum value of A be q and the maximum - p. \[\frac{{az^2+c}}{{az^2-bz+c}} \geq q \Rightarrow q > 0 \Rightarrow 0 \geq a(q-1)z^2 - bqz + c(q-1) \Rightarrow q \leq 1 \text{ and } D=0 \Rightarrow \\ b^2q^2 = 4ac(q-1)^2 \Rightarrow \]\[bq = \pm\ 2\sqrt{ac}(q-1) \Rightarrow bq = -2\sqrt{ac}(q-1) \text{ (because } 0 < q \leq 1) \Rightarrow q = \frac{{2\sqrt{ac}}}{{2\sqrt{2ac}+b}}\]We do the same for p and get that p$\geq$1 and p= $\frac{{2\sqrt{ac}}}{{2\sqrt{2ac}-b}}$ And we get that: $\boxed{\frac{{2d\sqrt{ac}}}{{2\sqrt{ac}-b}} \geq f(x,y) \geq \frac{{2d\sqrt{ac}}}{{2\sqrt{ac}+b}}}$