Given:
\[
X \sim \text{Uniform}(-1,1)
\]
\[
Y \mid X=x \sim \text{Uniform}(x^2 - 0.1,\; x^2 + 0.1)
\]
Step 1: Compute conditional expectation
For a uniform distribution on \([a,b]\), the mean is
\[
\frac{a+b}{2}.
\]
Thus,
\[
E[Y \mid X=x]
=
\frac{(x^2 - 0.1) + (x^2 + 0.1)}{2}
=
x^2.
\]
Hence,
\[
E[Y \mid X] = X^2.
\]
Step 2: Compute covariance
We use
\[
\mathrm{Cov}(X,Y) = E[XY] - E[X]E[Y].
\]
To compute \(E[XY]\), we use the Law of Total Expectation
(Tower Property):
\[
E[XY] = E\big[E[XY \mid X]\big].
\]
Since we are conditioning on X, the variable X becomes constant inside the conditional expectation.
Therefore,
\[
E[XY \mid X] = X E[Y \mid X].
\]
Thus,
\[
E[XY] = E\big[X E[Y \mid X]\big].
\]
Since \(E[Y \mid X] = X^2\),
\[
E[XY] = E[X \cdot X^2] = E[X^3].
\]
Step 3: Use symmetry
Because
\[
X \sim \text{Uniform}(-1,1),
\]
the distribution is symmetric about 0.
Hence all odd moments are zero:
\[
E[X^3] = 0.
\]
Alternative Method to Compute \(E[X^3]\):
Since the density of \(X\) is
\[
f(x) = \frac{1}{2}, \quad -1 \le x \le 1,
\]
we compute directly:
\[
E[X^3]
=
\int_{-1}^{1} x^3 \cdot \frac{1}{2} \, dx
=
\frac{1}{2}
\int_{-1}^{1} x^3 \, dx.
\]
Because \(x^3\) is an odd function and the interval \([-1,1]\) is symmetric,
\[
\int_{-1}^{1} x^3 \, dx = 0.
\]
Therefore,
\[
E[X^3] = 0.
\]
Also,
\[
E[X] = 0.
\]
Therefore,
\[
\mathrm{Cov}(X,Y) = 0.
\]
Step 4: Compute correlation
\[
\mathrm{Corr}(X,Y)
=
\frac{\mathrm{Cov}(X,Y)}
{\sqrt{\mathrm{Var}(X)\mathrm{Var}(Y)}}.
\]
Since covariance is zero,
\[
\mathrm{Corr}(X,Y) = 0.
\]
Final Answer:
\[
\boxed{0}
\]
