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Let $H$ be the orthocenter of $A B C$ and $K, L, M$ be the feet of the perpendiculars from $A, B, C$ to their opposite sides of $A B C$. Also let $D$ be the intersection point of lines $B E$ and $C F$. From the power of a point, we have

$$B A \cdot B M = B C \cdot B K$$

and

$$C A \cdot C L = C B \cdot C K$$

Adding (1) and (2) we have:

$$C A \cdot C L + B A \cdot B M = B C \cdot B K + C B \cdot C K = B C(B K + C K) = B C^2$$

Combining (3) with the problem statement $B C^2 = B A \cdot B F + C E \cdot C A$ we have:

$$\begin{aligned} & B A \cdot B F - B A \cdot B M = C A \cdot C L - C E \cdot C A \\ & B A(B F - B M) = C A(C L - C E) \\ & B A \cdot F M = C A \cdot L E \\ & \quad \frac{L E}{F M} = \frac{A B}{A C} = \frac{B L}{C M} \end{aligned}$$

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Where the last equality follows from $\triangle A M C \sim \triangle A L B$. Now since $\frac{L E}{F M} = \frac{B L}{C M}$ and $\angle F M C = \angle E L B = 90^{\circ}$, we get that triangles $\triangle F M C \sim \triangle E L B$. From this similarity we get

$$\measuredangle A E D = \measuredangle A E B = \angle L E B = \measuredangle M F C = 180^{\circ} - \angle A F C = 180 - \angle A F D,$$

meaning points $A, D, E, F$ are concyclic.
Since both pairs $\{E, F\}$ and $\{M, L\}$ satisfy the problem condition, we must have this fixed point we are looking for is the second intersection of the circumcircles around $A F D E$ and $A M H L$. Let this point be $X$. We now prove that $X$ is fixed on the circumcircle of $A M H L$ (which would imply $X$ is fixed).
From the concyclicity we have

$$\angle X L E = 180^{\circ} - \measuredangle X L A = \measuredangle X M A = \angle X M F \text{ and } \measuredangle X E L = \measuredangle X E A = 180 - \measuredangle X F A = \measuredangle X F M$$

and from here we get $\triangle X L E \sim \triangle X M F$. This similarity gives us

$$\frac{X L}{X M} = \frac{L E}{M F}.$$

Now combining (4) and (5) we get $\frac{X L}{X M} = \frac{A B}{A C}$ which is a fixed quantity. Since points $M, L$, the circumcircle of $A M L$, and ratio $\frac{X L}{X M}$ are fixed, this implies that point $X$ is fixed.

1. Given Condition and Setup:
We are given an acute triangle $ABC$ with variable points $E$ and $F$ on sides $AC$ and $AB$ respectively such that:
$$BC^2 = BA \cdot BF + CE \cdot CA$$
We need to prove that the circumcircle of $\triangle AEF$ passes through a fixed point other than $A$.

2. Intersection of Circles:
Consider the circumcircles of $\triangle AEB$ and $\triangle AFC$. Let these circles intersect $BC$ at some point $D$. This point $D$ is crucial as it helps in defining the positions of $E$ and $F$.

3. **Inversion about Point $A$:**
Perform an inversion about point $A$ with an arbitrary radius. Under this inversion, points $B$, $C$, $E$, and $F$ map to points $B'$, $C'$, $E'$, and $F'$ respectively. The inversion transforms the circumcircles of $\triangle AEB$ and $\triangle AFC$ into lines passing through $A'$, the image of $A$.

4. Cyclic Quadrilateral:
The inversion transforms the cyclic quadrilateral $ABDC$ into another cyclic quadrilateral $A'B'D'C'$. Here, $A'$, $B'$, and $C'$ are fixed points, while $D'$ is the image of the intersection point $D$ and is variable.

5. Intersection Points:
The points $E'$ and $F'$ are defined as:
$$E' = A'B' \cap C'D' \quad \text{and} \quad F' = A'C' \cap B'D'$$
We need to show that the line $E'F'$ passes through a fixed point.

6. **Pole of Line $B'C'$:**
Let $P$ be the pole of the line $B'C'$ with respect to the circumcircle of $\triangle A'B'C'$. The pole $P$ is a fixed point because it depends only on the fixed points $A'$, $B'$, and $C'$.

7. **Line $E'F'$ Passing Through $P$:**
It is a well-known result in projective geometry that the line $E'F'$ passes through the pole $P$ of the line $B'C'$ with respect to the circumcircle of $\triangle A'B'C'$. Therefore, $P$ is the fixed point through which the line $E'F'$ always passes.

8. Conclusion:
Since the line $E'F'$ always passes through the fixed point $P$, the original circumcircle of $\triangle AEF$ must pass through the fixed point corresponding to $P$ under the inversion.

$\blacksquare$