Answer
The partial derivatives $f_{x}$ and $f_{y}$ exist near (1,1), and are continuous functions so $f$ is differentiable at (1,1).
(by Theorem 8)
$L(x, y)=3x+4y-6$
Work Step by Step
$f(x, y)=x^{3}y^{4}$.
$f_{x}(x, y)=3x^{2}y^{4}$ and $f_{y}(x, y)=4x^{3}y^{3}$.
The partial derivatives $f_{x}$ and $f_{y}$ exist near (1,1),
and are continuous functions so $f$ is differentiable at (1,1).
(by Theorem 8)
The linearization of $f$ at (a,b):
$f(x, y)\approx L(x, y)=f(a, b)+ dz$, where
$dz=f_{x}(a, b)(x-a)+f_{y}(a, b)(y-b)$
$f_{x}(1,1)=3$,
$f_{y}(1,1)=4$,
$L(x, y)=f(1,1)+f_{x}(1,1)(x-1)+f_{y}(1,1)(y-1)$
$L(x, y)=1+3(x-1)+4(y-1)$
$L(x, y)=1+3x-3+4y-4$
$L(x, y)=3x+4y-6$.
