Find an equation of the tangent plane to the surface $$(u,v)=\langle u^2+v^2,2uv,u^3\rangle$$at the point$$(5,4,1)$$ i know the tangent equation, you need the partial of x and y.. plug in the points. plug back into equation and bam, easy and done. but thats when you are given a normal function... why does this look so confusing? do i have to calculate something else?
edit: I am not sure if i am over thinking it, or if what i did is correct. please help clarify $$ f_u=2u+2v+3u^2 $$ $$f_v=2v+2u$$plugged 5 and 4 in (for u and v) $$\langle 10,8,75\rangle\langle 8,10,0\rangle$$ took the cross product to find a vector.. and thats what i used in the tangent equation. Did i go about this correctly..?
As the first step you need to fix an error in the first line of your question: $$(u,v)=\langle u^2+v^2,2uv,u^3\rangle$$ You can not make a pair of numbers $(u,v)$ equal to a triple $(u^2+v^2, 2uv, u^3)$.
Your given surface is defined as a set of all points in 3D space whose coordinates fit given formulas (assume it is Euclidean space with rectangular Cartesian coordinates' system). The surface itself is a two-dimensional object so it is described with two parameters $u$ and $v$, which denote some 'directions' of walking on the surface. Every possible pair $(u,v)$ of those two parameters values identify some point $P$ on the surface, so either you should say 'we have a point $P$, whose position depending on $u$ and $v$ is such and such': $$P(u,v) = (u^2+v^2, 2uv, u^3)$$ or 'we have a function which assigns a point to every pair of parameters in such a way': $$(u,v) \mapsto (u^2+v^2, 2uv, u^3)$$
Now, you need to identify parameters values for the given point: $$(u^2+v^2, 2uv, u^3) = (5, 4, 1)$$ This makes a system of equations $$\begin{cases} u^2+v^2 & = 5 \\ 2uv & = 4 \\ u^3 & = 1 \end{cases}$$ which has a unique solution: $$(u, v) = (1, 2)$$
To determine a plane tangent to the surface in the point, we find two lines tangent to the surface first. The lines are found by testing in what directions will the point $P(u,v)$ move in our 3D-space from the given point with infinitesimal change of the parameters. One direction is given by a derivative $\frac{\partial P}{\partial u}$ and another one with $\frac{\partial P}{\partial v}$: $$P_u(u,v) = \frac\partial {\partial u} (u^2+v^2, 2uv, u^3) = (2u, 2v, 3u^2)$$ $$P_v(u,v) = \frac\partial {\partial v} (u^2+v^2, 2uv, u^3) = (2v, 2u, 0)$$
Substitute the given point $(u,v)=(1,2)$ and we get two vectors: $$P_u(1,2) = (2, 4, 3)$$ $$P_v(1,2) = (4, 2, 0)$$ tangent to the surface. These vectors define two lines through $P(1,2) = (5,4,1)$, tangent to the surface, and those lines define a plane you seek.
Cross-multiply them to get a new vector $N$, orthogonal to both of them. That is a direction vector for a plane: $$N(1,2) = P_u(1,2) \times P_v(1,2) = (2, 4, 3)\times(4, 2, 0) = (-6, 12, -12)$$ We can shorten it, as we're interested in its direction, not a length. So let's replace $N$ with $n = N/6 = (-1, 2, -2)$.
Now a general equation of a plane orthogonal to vector $n$ is $$n \cdot position = \mathrm{const.}$$ (note the dot is a scalar product of vectors here!) or expanded: $$n_x\cdot x+n_y\cdot y+n_z\cdot z=D$$ for some constant $D$.
We have $n = N/6 = (-1, 2, -2)$, so the equation is $$-x + 2 y - 2 z=D$$ The last step is finding such $D$ which makes the plane to pass $(5,4,1)$. We do it by substituting the point to the equation: $$-5 + 2\cdot 4 - 2\cdot 1=D$$ and we find $$D = 1$$
Finally the equation of the plane is $$-x + 2 y - 2 z = 1$$ or equivalent $$x - 2 y + 2 z + 1 = 0$$