Find $f_{x}(x, y)$ and $f_{y}(x, y)$. Then find $f_{x}(2,-1)$ and $f_{y}(-2,-1)$. \[ f(x, y)=-5 e^{6 x-5 y} \] \[ f_{x}(x, y)= \]

Step 1 ：First, we need to find the partial derivatives $f_{x}(x, y)$ and $f_{y}(x, y)$ of the function $f(x, y)=-5 e^{6 x-5 y}$.

Step 2 ：The partial derivative of $f(x, y)$ with respect to $x$ is given by $f_{x}(x, y) = \frac{\partial}{\partial x}(-5 e^{6 x-5 y})$.

Step 3 ：Using the chain rule, we get $f_{x}(x, y) = -5 \cdot 6 e^{6 x-5 y} = -30 e^{6 x-5 y}$.

Step 4 ：Similarly, the partial derivative of $f(x, y)$ with respect to $y$ is given by $f_{y}(x, y) = \frac{\partial}{\partial y}(-5 e^{6 x-5 y})$.

Step 5 ：Using the chain rule again, we get $f_{y}(x, y) = -5 \cdot -5 e^{6 x-5 y} = 25 e^{6 x-5 y}$.

Step 6 ：Now, we can find $f_{x}(2,-1)$ by substituting $x=2$ and $y=-1$ into $f_{x}(x, y)$.

Step 7 ：We get $f_{x}(2,-1) = -30 e^{6 \cdot 2 - 5 \cdot -1} = -30 e^{17}$.

Step 8 ：Similarly, we can find $f_{y}(-2,-1)$ by substituting $x=-2$ and $y=-1$ into $f_{y}(x, y)$.

Step 9 ：We get $f_{y}(-2,-1) = 25 e^{6 \cdot -2 - 5 \cdot -1} = 25 e^{-17}$.

Step 10 ：So, the final answers are $f_{x}(2,-1) = -30 e^{17}$ and $f_{y}(-2,-1) = 25 e^{-17}$.