How would I go about finding the tangent plane to the equation $(cosh x_1)^2 = x_2^2+x_3^2+x_4^2$ at the point $(0,0,\frac 1{\sqrt 2},\frac 1{\sqrt 2})$?
I tried to find it by having $f=-(cosh x_1)^2+x_2^2+x_3^2+x_4^2$ and then finding the gradient of $f$.
Then I set up the equation of the plane as $a(x_1-0)+b(x_2-0)+c(x_3-\frac 1{\sqrt 2})+d(x_4-\frac 1{\sqrt 2})=0$ where $a,b,c,d$ are the partial derivatives of $f$ for $x_1,x_2,x_3,x_4$ respectively. After doing the plugging and chugging my final equation for the plane was $\frac 2{\sqrt 2}x_3+\frac 2{\sqrt 2}x_4=0$ which just doesn't look quite right.
Also, once I find the equation of the plane, how could I put the equation in the form of $(0,0,\frac 1{2^{1/2}},\frac 1{2^{1/2}})+span(v_1,v_2,v_3)$?
Your method seems sound, and I agree with you on the values of $a$, $b$, $c$, and $d$. But what doesn't “look right” about your hyperplane is that it doesn't pass through $\left(0,0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$.
Your normal form of the equation for the hyperplane needs a nonzero parameter on the right-hand side. You can find the value of that parameter by evaluating the left-hand side at $\left(0,0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$. So: $$ \frac{2}{\sqrt{2}} x_3 + \frac{2}{\sqrt{2}} x_4 = \frac{2}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + \frac{2}{\sqrt{2}}\cdot \frac{1}{\sqrt{2}} \implies x_3 + x_4 = \sqrt{2} $$ Your $v_1$, $v_2$, and $v_3$ are any three linearly independent vectors whose third and fourth components add to zero. They span the nullspace of the $1\times 4$ matrix $$ \begin{bmatrix} a & b & c & d \end{bmatrix} = \begin{bmatrix} 0 & 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix} \longrightarrow \begin{bmatrix} 0 & 0 & 1 & 1 \end{bmatrix} $$ In linear algebra parlance, $x_1$, $x_2$, and $x_4$ are free variables, while $x_3 = -x_4$. We get one vector from setting each of the free variables equal to $1$ and the others to zero, one at a time. If $x_1 = 1$ and $x_2 = x_4 = 0$, we get $v_1 = (1,0,0,0)$. If $x_2 = 1$ and $x_1 = x_4 = 0$, we get $v_2 = (0,1,0,0)$. If $x_4 = 1$ and $x_1 = x_2 = 0$, we get $v_3 = (0,0,-1,1)$.