Find the minimum and maximum value of the function 
\begin{align*} f(x,y)=ax^2+cy^2 \end{align*}
Under the condition $ax^2-bxy+cy^2=d$, where $a,b,c,d$ are positive real numbers such that $b^2 -4ac <0$
1. We start with the given function \( f(x,y) = ax^2 + cy^2 \) and the constraint \( ax^2 - bxy + cy^2 = d \), where \( a, b, c, d \) are positive real numbers and \( b^2 - 4ac < 0 \).

2. To simplify the problem, we introduce the variable \( z = \frac{x}{y} \). This allows us to rewrite the constraint in terms of \( z \):
   \[
   ax^2 - bxy + cy^2 = d \implies a\left(\frac{x}{y}\right)^2 y^2 - b\left(\frac{x}{y}\right)y^2 + cy^2 = d \implies az^2 y^2 - bzy^2 + cy^2 = d
   \]
   Dividing through by \( y^2 \), we get:
   \[
   az^2 - bz + c = \frac{d}{y^2} \implies y^2 = \frac{d}{az^2 - bz + c}
   \]

3. Similarly, we can express \( x^2 \) in terms of \( z \):
   \[
   ax^2 - bxy + cy^2 = d \implies a - \frac{b}{z} + \frac{c}{z^2} = \frac{d}{x^2} \implies x^2 = \frac{dz^2}{az^2 - bz + c}
   \]

4. Now, we express \( f(x,y) \) in terms of \( z \):
   \[
   f(x,y) = ax^2 + cy^2 = a \left( \frac{dz^2}{az^2 - bz + c} \right) + c \left( \frac{d}{az^2 - bz + c} \right) = \frac{d(az^2 + c)}{az^2 - bz + c}
   \]
   Let \( A = \frac{az^2 + c}{az^2 - bz + c} \). We need to find the minimum and maximum values of \( A \).

5. To find the minimum value of \( A \), we set up the inequality:
   \[
   \frac{az^2 + c}{az^2 - bz + c} \geq q \implies q(az^2 - bz + c) \leq az^2 + c
   \]
   Simplifying, we get:
   \[
   qaz^2 - qbz + qc \leq az^2 + c \implies (qa - a)z^2 - qbz + (qc - c) \leq 0 \implies a(q-1)z^2 - bqz + c(q-1) \leq 0
   \]
   For this quadratic inequality to hold for all \( z \), the discriminant must be non-positive:
   \[
   (bq)^2 - 4a(q-1)c(q-1) \leq 0 \implies b^2q^2 \leq 4ac(q-1)^2
   \]
   Solving for \( q \), we get:
   \[
   bq = \pm 2\sqrt{ac}(q-1)
   \]
   Since \( 0 < q \leq 1 \), we take the negative root:
   \[
   bq = -2\sqrt{ac}(q-1) \implies bq = -2\sqrt{ac}q + 2\sqrt{ac} \implies q(b + 2\sqrt{ac}) = 2\sqrt{ac} \implies q = \frac{2\sqrt{ac}}{b + 2\sqrt{ac}}
   \]

6. To find the maximum value of \( A \), we set up the inequality:
   \[
   \frac{az^2 + c}{az^2 - bz + c} \leq p \implies p(az^2 - bz + c) \geq az^2 + c
   \]
   Simplifying, we get:
   \[
   paz^2 - pbz + pc \geq az^2 + c \implies (pa - a)z^2 - pbz + (pc - c) \geq 0 \implies a(p-1)z^2 - pbz + c(p-1) \geq 0
   \]
   For this quadratic inequality to hold for all \( z \), the discriminant must be non-negative:
   \[
   (pb)^2 - 4a(p-1)c(p-1) \geq 0 \implies b^2p^2 \geq 4ac(p-1)^2
   \]
   Solving for \( p \), we get:
   \[
   bp = \pm 2\sqrt{ac}(p-1)
   \]
   Since \( p \geq 1 \), we take the positive root:
   \[
   bp = 2\sqrt{ac}(p-1) \implies bp = 2\sqrt{ac}p - 2\sqrt{ac} \implies p(b - 2\sqrt{ac}) = 2\sqrt{ac} \implies p = \frac{2\sqrt{ac}}{b - 2\sqrt{ac}}
   \]

7. Therefore, the minimum and maximum values of \( f(x,y) \) are:
   \[
   \boxed{\frac{2d\sqrt{ac}}{b + 2\sqrt{ac}} \leq f(x,y) \leq \frac{2d\sqrt{ac}}{b - 2\sqrt{ac}}}
   \]