Find dy/dx and d2y/dx2 at (1,1) for the equation ((x-h)2)/(a2) + ((y-k)2)/(b2) = 1

The equation is:

$$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$$

Differentiating both sides with respect to $x$:

$$\frac{2(x-h)}{a^2} + \frac{2(y-k)}{b^2} \frac{dy}{dx} = 0$$

$$\frac{2(y-k)}{b^2} \frac{dy}{dx} = -\frac{2(x-h)}{a^2}$$ $$\frac{dy}{dx} = -\frac{2(x-h)}{a^2} \cdot \frac{b^2}{2(y-k)}$$ $$\frac{dy}{dx} = -\frac{b^2(x-h)}{a^2(y-k)}$$

Substitute $x=1$ and $y=1$ into the expression for $\frac{dy}{dx}$:

$$\frac{dy}{dx} \Big|_{(1,1)} = -\frac{b^2(1-h)}{a^2(1-k)}$$

The first derivative at $(1,1)$ is $\boxed{-\frac{b^2(1-h)}{a^2(1-k)}}$.

We have $\frac{dy}{dx} = -\frac{b^2}{a^2} \frac{x-h}{y-k}$.
Using the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$, where $u = x-h$ and $v = y-k$:
$u' = 1$
$v' = \frac{dy}{dx}$

$$\frac{d^2y}{dx^2} = -\frac{b^2}{a^2} \frac{1 \cdot (y-k) - (x-h) \frac{dy}{dx}}{(y-k)^2}$$

Substitute $\frac{dy}{dx} = -\frac{b^2(x-h)}{a^2(y-k)}$:

$$\frac{d^2y}{dx^2} = -\frac{b^2}{a^2} \frac{(y-k) - (x-h) \left(-\frac{b^2(x-h)}{a^2(y-k)}\right)}{(y-k)^2}$$ $$\frac{d^2y}{dx^2} = -\frac{b^2}{a^2} \frac{(y-k) + \frac{b^2(x-h)^2}{a^2(y-k)}}{(y-k)^2}$$

To simplify the numerator, find a common denominator:

$$\frac{d^2y}{dx^2} = -\frac{b^2}{a^2} \frac{\frac{a^2(y-k)^2 + b^2(x-h)^2}{a^2(y-k)}}{(y-k)^2}$$ $$\frac{d^2y}{dx^2} = -\frac{b^2}{a^4} \frac{a^2(y-k)^2 + b^2(x-h)^2}{(y-k)^3}$$

From the original equation, $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$.
Multiply by $a^2 b^2$: $b^2(x-h)^2 + a^2(y-k)^2 = a^2 b^2$.
Substitute this into the expression for $\frac{d^2y}{dx^2}$:

$$\frac{d^2y}{dx^2} = -\frac{b^2}{a^4} \frac{a^2 b^2}{(y-k)^3}$$ $$\frac{d^2y}{dx^2} = -\frac{b^4}{a^2(y-k)^3}$$

Substitute $y=1$ into the expression for $\frac{d^2y}{dx^2}$:

$$\frac{d^2y}{dx^2} \Big|_{(1,1)} = -\frac{b^4}{a^2(1-k)^3}$$

The second derivative at $(1,1)$ is $\boxed{-\frac{b^4}{a^2(1-k)^3}}$.