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Work where $x\ne 0$ and $f$ is differentiable. Set $q=y/x$. The chain rule givesConsequentlyGeometrically, a radial rescaling of both coordinates leaves $y/x$ unchanged, so the function has zero rate of change along radial dilations. No value or derivative at $x=0$ is implied by this representation.

Let $h=x-\pi /3$. At the expansion point, the function and its first three derivatives have values $1/2$, $-\sqrt{3}/2$, $-1/2$ and $\sqrt{3}/2$. Taylor's theorem therefore givesFor real $x$, the magnitude of the omitted fourth-order remainder is at most $|h{|}^4/24$, because the fourth derivative is another cosine and has magnitude at most one.

The gradient is $\nabla f=(8x,2y)$, so $\nabla f(1,1)=(8,2)$. The supplied direction is not a unit vector; normalize it to $\hat{\mathbf{u}}=(1,2)/\sqrt{5}$. Its directional derivative isThe direction of greatest decrease is opposite to the gradient:The Cauchy-Schwarz inequality shows that no other unit direction can give a more negative derivative.

Write the plane as $z=6-3x-2y$. Its projection on the $xy$ plane is the triangle $0\le x\le 2$, $0\le y\le 3-3x/2$. The scalar surface area element isThus the weighted surface integral isA geometric check uses the triangular face's area $3\sqrt{14}$ and mean $y$ coordinate $(0+3+0)/3=1$. Multiplying them gives the same result, because $y$ is an affine function on this planar triangle. Its edges have zero surface measure, so strict or non-strict octant inequalities do not change the integral.

Substitute the path into the height before differentiating: $x^2=1+t$, $y=t-1$, soThe ordinary chain rule immediately yieldsThe multivariable calculation gives the same result: $h_x=-2x/(x^2+2y^4+1{)}^2$, $h_y=-8y^3/(x^2+2y^4+1{)}^2$, and $\dot{h}=h_x\dot{x}+h_y\dot{y}$ with $\dot{x}=1/(2\sqrt{1+t})$ and $\dot{y}=1$. The real path has $t\ge -1$; this chain-rule calculation applies for $t>-1$, and the simplified height has a finite right derivative at the endpoint even though $\dot{x}$ there is singular.

Use the outward normal vector. The vector field is $\mathbf{F}=(xz,2y,-z^2)$, with divergence $\nabla \cdot \mathbf{F}=z+2-2z=2-z$. By the divergence theorem, its outward flux equals the integral of this divergence over the solid cylinder. Each horizontal slice has area $8\pi$, soFor an independent boundary check, the cylindrical side contributes $84\pi$, the top contributes $(-3^2)(8\pi )=-72\pi$, and the bottom contributes zero. Their sum is again $12\pi$. Reversing all normals changes its sign.