Let the isosceles right triangle $ABC$ with  $\angle A= 90^o$ . The points  $E$ and $F$ are taken on the ray $AC$ so that $\angle ABE   = 15^o$ and $CE = CF$. Determine the measure of the angle $CBF$.
1. Given an isosceles right triangle \( \triangle ABC \) with \( \angle A = 90^\circ \). Points \( E \) and \( F \) are on the ray \( AC \) such that \( \angle ABE = 15^\circ \) and \( CE = CF \).

2. Let \( CE = CF = a \).

3. Extend the diagram by adding point \( D \) such that \( DE = CE = a \) and \( \angle DEC = 60^\circ \).

4. By simple angle chasing, we find \( \angle BED = 45^\circ \).

5. Using the Law of Sines in \( \triangle BED \):
   \[
   \frac{BE}{\sin 30^\circ} = \frac{a}{\sin 45^\circ}
   \]
   Solving for \( BE \):
   \[
   BE = a \cdot \frac{\sin 45^\circ}{\sin 30^\circ} = a \cdot \frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}} = a \sqrt{2}
   \]

6. Using the Law of Cosines in \( \triangle BDE \):
   \[
   BD^2 = DE^2 + BE^2 - 2 \cdot DE \cdot BE \cdot \cos 45^\circ
   \]
   Substituting \( DE = a \) and \( BE = a \sqrt{2} \):
   \[
   BD^2 = a^2 + (a \sqrt{2})^2 - 2 \cdot a \cdot a \sqrt{2} \cdot \frac{\sqrt{2}}{2}
   \]
   Simplifying:
   \[
   BD^2 = a^2 + 2a^2 - 2a^2 = a^2
   \]
   Thus:
   \[
   BD = a
   \]

7. Since \( BD = DE \), \( \triangle BDE \) is isosceles with \( \angle BED = 45^\circ \). Therefore, \( \angle BDE = \angle BED = 45^\circ \).

8. Since \( \angle DEC = 60^\circ \), we have:
   \[
   \angle BDC = 180^\circ - \angle BDE - \angle DEC = 180^\circ - 45^\circ - 60^\circ = 75^\circ
   \]

9. Since \( \angle BCD = 15^\circ \) and \( \angle CDF = 30^\circ \), points \( B \), \( D \), and \( F \) are collinear.

10. Therefore, \( \angle CBF = \angle CBD = 15^\circ \).

The final answer is \(\boxed{15^\circ}\).