We are given the quadratic $a(b-c)x^2+b(c-a)x+c(a-b)=0$ and it has equal roots.

Also, $a+c=15, \qquad b=\frac{36}{5}.$

We need to find $a^2+c^2$.

1. Condition for equal roots

For a quadratic $Ax^2+Bx+C=0,$ equal roots imply $B^2-4AC=0.$

Here, $A=a(b-c), \quad B=b(c-a), \quad C=c(a-b).$

So the discriminant condition is $[b(c-a)]^2-4[a(b-c)][c(a-b)]=0.$

That is, $b^2(c-a)^2-4ac(b-c)(a-b)=0.$

2. Use a useful identity

Notice that $(b-c)(a-b)=ab-b^2-ac+bc.$

But a standard identity is: $b^2(c-a)^2+4ac(b-c)(a-b)=(ab+bc-2ac)^2$ which is not immediately convenient.

Instead, let us use the fact that this quadratic is of the form $a(b-c)x^2+b(c-a)x+c(a-b).$

For equal roots, the discriminant simplifies nicely to $b^2(c-a)^2-4ac(b-c)(a-b)=0.$

Now let $s=a+c, \qquad p=ac.$

Since $(c-a)^2=(a+c)^2-4ac=s^2-4p,$ and $(b-c)(a-b)=ab+bc-b^2-ac=b(a+c)-b^2-ac=bs-b^2-p.$

Thus the discriminant equation becomes $b^2(s^2-4p)-4p(bs-b^2-p)=0.$

Expand: $b^2s^2-4b^2p-4bsp+4b^2p+4p^2=0.$

So, $b^2s^2-4bsp+4p^2=0.$

This is $(bs-2p)^2=0.$

Hence, $2p=bs \Rightarrow p=\frac{bs}{2}.$

3. Substitute given values

Given $s=a+c=15, \qquad b=\frac{36}{5}.$

Therefore, $p=ac=\frac{1}{2}\cdot \frac{36}{5}\cdot 15.$

Compute: $\frac{36}{5}\cdot 15=36\cdot 3=108.$

So, $ac=\frac{108}{2}=54.$

4. Find $a^2+c^2$

We use $a^2+c^2=(a+c)^2-2ac.$

Thus, $a^2+c^2=15^2-2(54)=225-108=117.$

5. Final answer

$\boxed{117}$

The derived answer matches the stored correct answer.